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# 4.6 Exponential and logarithmic equations  (Page 7/8)

 Page 7 / 8

When does an extraneous solution occur? How can an extraneous solution be recognized?

When can the one-to-one property of logarithms be used to solve an equation? When can it not be used?

The one-to-one property can be used if both sides of the equation can be rewritten as a single logarithm with the same base. If so, the arguments can be set equal to each other, and the resulting equation can be solved algebraically. The one-to-one property cannot be used when each side of the equation cannot be rewritten as a single logarithm with the same base.

## Algebraic

For the following exercises, use like bases to solve the exponential equation.

${4}^{-3v-2}={4}^{-v}$

$64\cdot {4}^{3x}=16$

$x=-\frac{1}{3}$

${3}^{2x+1}\cdot {3}^{x}=243$

${2}^{-3n}\cdot \frac{1}{4}={2}^{n+2}$

$n=-1$

$625\cdot {5}^{3x+3}=125$

$\frac{{36}^{3b}}{{36}^{2b}}={216}^{2-b}$

$b=\frac{6}{5}$

${\left(\frac{1}{64}\right)}^{3n}\cdot 8={2}^{6}$

For the following exercises, use logarithms to solve.

${9}^{x-10}=1$

$x=10$

$2{e}^{6x}=13$

${e}^{r+10}-10=-42$

No solution

$2\cdot {10}^{9a}=29$

$-8\cdot {10}^{p+7}-7=-24$

$p=\mathrm{log}\left(\frac{17}{8}\right)-7$

$7{e}^{3n-5}+5=-89$

${e}^{-3k}+6=44$

$k=-\frac{\mathrm{ln}\left(38\right)}{3}$

$-5{e}^{9x-8}-8=-62$

$-6{e}^{9x+8}+2=-74$

$x=\frac{\mathrm{ln}\left(\frac{38}{3}\right)-8}{9}$

${2}^{x+1}={5}^{2x-1}$

${e}^{2x}-{e}^{x}-132=0$

$7{e}^{8x+8}-5=-95$

$10{e}^{8x+3}+2=8$

$x=\frac{\mathrm{ln}\left(\frac{3}{5}\right)-3}{8}$

$4{e}^{3x+3}-7=53$

$8{e}^{-5x-2}-4=-90$

no solution

${3}^{2x+1}={7}^{x-2}$

${e}^{2x}-{e}^{x}-6=0$

$x=\mathrm{ln}\left(3\right)$

$3{e}^{3-3x}+6=-31$

For the following exercises, use the definition of a logarithm to rewrite the equation as an exponential equation.

$\mathrm{log}\left(\frac{1}{100}\right)=-2$

${10}^{-2}=\frac{1}{100}$

${\mathrm{log}}_{324}\left(18\right)=\frac{1}{2}$

For the following exercises, use the definition of a logarithm to solve the equation.

$5{\mathrm{log}}_{7}n=10$

$n=49$

$-8{\mathrm{log}}_{9}x=16$

$4+{\mathrm{log}}_{2}\left(9k\right)=2$

$k=\frac{1}{36}$

$2\mathrm{log}\left(8n+4\right)+6=10$

$10-4\mathrm{ln}\left(9-8x\right)=6$

$x=\frac{9-e}{8}$

For the following exercises, use the one-to-one property of logarithms to solve.

$\mathrm{ln}\left(10-3x\right)=\mathrm{ln}\left(-4x\right)$

${\mathrm{log}}_{13}\left(5n-2\right)={\mathrm{log}}_{13}\left(8-5n\right)$

$n=1$

$\mathrm{log}\left(x+3\right)-\mathrm{log}\left(x\right)=\mathrm{log}\left(74\right)$

$\mathrm{ln}\left(-3x\right)=\mathrm{ln}\left({x}^{2}-6x\right)$

No solution

${\mathrm{log}}_{4}\left(6-m\right)={\mathrm{log}}_{4}3m$

$\mathrm{ln}\left(x-2\right)-\mathrm{ln}\left(x\right)=\mathrm{ln}\left(54\right)$

No solution

${\mathrm{log}}_{9}\left(2{n}^{2}-14n\right)={\mathrm{log}}_{9}\left(-45+{n}^{2}\right)$

$\mathrm{ln}\left({x}^{2}-10\right)+\mathrm{ln}\left(9\right)=\mathrm{ln}\left(10\right)$

$x=±\frac{10}{3}$

For the following exercises, solve each equation for $\text{\hspace{0.17em}}x.$

$\mathrm{log}\left(x+12\right)=\mathrm{log}\left(x\right)+\mathrm{log}\left(12\right)$

$\mathrm{ln}\left(x\right)+\mathrm{ln}\left(x-3\right)=\mathrm{ln}\left(7x\right)$

$x=10$

${\mathrm{log}}_{2}\left(7x+6\right)=3$

$\mathrm{ln}\left(7\right)+\mathrm{ln}\left(2-4{x}^{2}\right)=\mathrm{ln}\left(14\right)$

$x=0$

${\mathrm{log}}_{8}\left(x+6\right)-{\mathrm{log}}_{8}\left(x\right)={\mathrm{log}}_{8}\left(58\right)$

$\mathrm{ln}\left(3\right)-\mathrm{ln}\left(3-3x\right)=\mathrm{ln}\left(4\right)$

$x=\frac{3}{4}$

${\mathrm{log}}_{3}\left(3x\right)-{\mathrm{log}}_{3}\left(6\right)={\mathrm{log}}_{3}\left(77\right)$

## Graphical

For the following exercises, solve the equation for $\text{\hspace{0.17em}}x,$ if there is a solution . Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution.

${\mathrm{log}}_{9}\left(x\right)-5=-4$

$x=9$

${\mathrm{log}}_{3}\left(x\right)+3=2$

$\mathrm{ln}\left(3x\right)=2$

$x=\frac{{e}^{2}}{3}\approx 2.5$

$\mathrm{ln}\left(x-5\right)=1$

$\mathrm{log}\left(4\right)+\mathrm{log}\left(-5x\right)=2$

$x=-5$

$-7+{\mathrm{log}}_{3}\left(4-x\right)=-6$

$\mathrm{ln}\left(4x-10\right)-6=-5$

$x=\frac{e+10}{4}\approx 3.2$

$\mathrm{log}\left(4-2x\right)=\mathrm{log}\left(-4x\right)$

${\mathrm{log}}_{11}\left(-2{x}^{2}-7x\right)={\mathrm{log}}_{11}\left(x-2\right)$

No solution

$\mathrm{ln}\left(2x+9\right)=\mathrm{ln}\left(-5x\right)$

${\mathrm{log}}_{9}\left(3-x\right)={\mathrm{log}}_{9}\left(4x-8\right)$

$x=\frac{11}{5}\approx 2.2$

$\mathrm{log}\left({x}^{2}+13\right)=\mathrm{log}\left(7x+3\right)$

$\frac{3}{{\mathrm{log}}_{2}\left(10\right)}-\mathrm{log}\left(x-9\right)=\mathrm{log}\left(44\right)$

$x=\frac{101}{11}\approx 9.2$

$\mathrm{ln}\left(x\right)-\mathrm{ln}\left(x+3\right)=\mathrm{ln}\left(6\right)$

For the following exercises, solve for the indicated value, and graph the situation showing the solution point.

An account with an initial deposit of $\text{\hspace{0.17em}}\text{6,500}\text{\hspace{0.17em}}$ earns $\text{\hspace{0.17em}}7.25%\text{\hspace{0.17em}}$ annual interest, compounded continuously. How much will the account be worth after 20 years?

about $\text{\hspace{0.17em}}27,710.24$

The formula for measuring sound intensity in decibels $\text{\hspace{0.17em}}D\text{\hspace{0.17em}}$ is defined by the equation $\text{\hspace{0.17em}}D=10\mathrm{log}\left(\frac{I}{{I}_{0}}\right),\text{}$ where $\text{\hspace{0.17em}}I\text{\hspace{0.17em}}$ is the intensity of the sound in watts per square meter and $\text{\hspace{0.17em}}{I}_{0}={10}^{-12}\text{\hspace{0.17em}}$ is the lowest level of sound that the average person can hear. How many decibels are emitted from a jet plane with a sound intensity of $\text{\hspace{0.17em}}8.3\cdot {10}^{2}\text{\hspace{0.17em}}$ watts per square meter?

The population of a small town is modeled by the equation $\text{\hspace{0.17em}}P=1650{e}^{0.5t}\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is measured in years. In approximately how many years will the town’s population reach $\text{\hspace{0.17em}}\text{20,000?}$

## Technology

For the following exercises, solve each equation by rewriting the exponential expression using the indicated logarithm. Then use a calculator to approximate $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ to 3 decimal places .

$1000{\left(1.03\right)}^{t}=5000\text{\hspace{0.17em}}$ using the common log.

${e}^{5x}=17\text{\hspace{0.17em}}$ using the natural log

$\frac{\mathrm{ln}\left(17\right)}{5}\approx 0.567$

$3{\left(1.04\right)}^{3t}=8\text{\hspace{0.17em}}$ using the common log

${3}^{4x-5}=38\text{\hspace{0.17em}}$ using the common log

$50{e}^{-0.12t}=10\text{\hspace{0.17em}}$ using the natural log

For the following exercises, use a calculator to solve the equation. Unless indicated otherwise, round all answers to the nearest ten-thousandth.

$7{e}^{3x-5}+7.9=47$

$x\approx 2.2401$

$\mathrm{ln}\left(3\right)+\mathrm{ln}\left(4.4x+6.8\right)=2$

$\mathrm{log}\left(-0.7x-9\right)=1+5\mathrm{log}\left(5\right)$

$x\approx -\text{44655}.\text{7143}$

Atmospheric pressure $\text{\hspace{0.17em}}P\text{\hspace{0.17em}}$ in pounds per square inch is represented by the formula $\text{\hspace{0.17em}}P=14.7{e}^{-0.21x},$ where $x$ is the number of miles above sea level. To the nearest foot, how high is the peak of a mountain with an atmospheric pressure of $\text{\hspace{0.17em}}8.369\text{\hspace{0.17em}}$ pounds per square inch? ( Hint : there are 5280 feet in a mile)

The magnitude M of an earthquake is represented by the equation $\text{\hspace{0.17em}}M=\frac{2}{3}\mathrm{log}\left(\frac{E}{{E}_{0}}\right)\text{\hspace{0.17em}}$ where $\text{\hspace{0.17em}}E\text{\hspace{0.17em}}$ is the amount of energy released by the earthquake in joules and $\text{\hspace{0.17em}}{E}_{0}={10}^{4.4}\text{\hspace{0.17em}}$ is the assigned minimal measure released by an earthquake. To the nearest hundredth, what would the magnitude be of an earthquake releasing $\text{\hspace{0.17em}}1.4\cdot {10}^{13}\text{\hspace{0.17em}}$ joules of energy?

about $\text{\hspace{0.17em}}5.83$

## Extensions

Use the definition of a logarithm along with the one-to-one property of logarithms to prove that $\text{\hspace{0.17em}}{b}^{{\mathrm{log}}_{b}x}=x.$

Recall the formula for continually compounding interest, $\text{\hspace{0.17em}}y=A{e}^{kt}.\text{\hspace{0.17em}}$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to a single logarithm.

$t=\mathrm{ln}\left({\left(\frac{y}{A}\right)}^{\frac{1}{k}}\right)$

Recall the compound interest formula $\text{\hspace{0.17em}}A=a{\left(1+\frac{r}{k}\right)}^{kt}.\text{\hspace{0.17em}}$ Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t.$

Newton’s Law of Cooling states that the temperature $\text{\hspace{0.17em}}T\text{\hspace{0.17em}}$ of an object at any time t can be described by the equation $\text{\hspace{0.17em}}T={T}_{s}+\left({T}_{0}-{T}_{s}\right){e}^{-kt},$ where $\text{\hspace{0.17em}}{T}_{s}\text{\hspace{0.17em}}$ is the temperature of the surrounding environment, $\text{\hspace{0.17em}}{T}_{0}\text{\hspace{0.17em}}$ is the initial temperature of the object, and $\text{\hspace{0.17em}}k\text{}$ is the cooling rate. Use the definition of a logarithm along with properties of logarithms to solve the formula for time $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ is equal to a single logarithm.

$t=\mathrm{ln}\left({\left(\frac{T-{T}_{s}}{{T}_{0}-{T}_{s}}\right)}^{-\text{\hspace{0.17em}}\frac{1}{k}}\right)$

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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
A hedge is contrusted to be in the shape of hyperbola near a fountain at the center of yard.the hedge will follow the asymptotes y=x and y=-x and closest distance near the distance to the centre fountain at 5 yards find the eqution of the hyperbola
A doctor prescribes 125 milligrams of a therapeutic drug that decays by about 30% each hour. To the nearest hour, what is the half-life of the drug?
Find the domain of the function in interval or inequality notation f(x)=4-9x+3x^2
hello
Outside temperatures over the course of a day can be modeled as a sinusoidal function. Suppose the high temperature of ?105°F??105°F? occurs at 5PM and the average temperature for the day is ?85°F.??85°F.? Find the temperature, to the nearest degree, at 9AM.
if you have the amplitude and the period and the phase shift ho would you know where to start and where to end?
rotation by 80 of (x^2/9)-(y^2/16)=1
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a²=4
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A bridge is to be built in the shape of a semi-elliptical arch and is to have a span of 120 feet. The height of the arch at a distance of 40 feet from the center is to be 8 feet. Find the height of the arch at its center


text:
# Upper bound on the divisorial gonality of a graph

$\rm{gon}(G) \leq \frac{|E(G)|-|V(G)|}{2} + 2$, where $\rm{gon}(G)$ the denotes the divisorial gonality of graph $G$.

## Remarks

• It is a conjecture of Matthew Baker based on extensive computer calculations by Adam Tart [1].
• In the context of graph divisor theory $|E(G)|-|V(G)|+ 1$ is called as the genus of the graph and denoted by $g$. Using this notion, the conjecture can be formulated as $\rm{gon}(G) \leq \lfloor \frac{g+3}{2} \rfloor$, which is the original form of the conjecture, see Conj. 3.10. in [1].
• It is also conjectured in [1] that for each integer $g \geq 0$, there exists a graph of genus $g$ with gonality exactly $\lfloor \frac{g+3}{2} \rfloor$.
• There is also a more general version of the conjecture in [1], the so-called Brill-Noether Conjecture for Graphs: Fix integers $g, r, d \geq 0$, and set $\varrho(g, r, d) = g − (r + 1)(g − d + r)$. Then:
• (1) If $\varrho(g, r, d) \geq 0$, then every graph of genus $g$ has a divisor $D$ with $\rm{rank}(D) = r$ and $\deg(D) \leq d$.
• (2) If $\varrho(g, r, d) \lt 0$, then there exists a graph of genus $g$ for which there is no divisor D with $\rm{rank}(D) = r$ and $\deg(D) \leq d$.

### References

1. M. Baker, Specialization of linear systems from curves to graphs, Algebra and Number Theory (2008) DOI link, Author Link

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# If the average (arithmetic mean) of a, b, c, and d is m, is their stan

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Math Revolution GMAT Instructor
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If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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01 Jan 2018, 01:20
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[GMAT math practice question]

If the average (arithmetic mean) of $$a, b, c$$, and $$d$$ is $$m$$, is their standard deviation greater than $$1$$?

1) $$a=1$$
2) $$m=4$$
[Reveal] Spoiler: OA

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If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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01 Jan 2018, 03:41
MathRevolution wrote:
[GMAT math practice question]

If the average (arithmetic mean) of $$a, b, c$$, and $$d$$ is $$m$$, is their standard deviation greater than $$1$$?

1) $$a=1$$
2) $$m=4$$

Statement 1: Nothing mentioned about other variables. if $$a=b=c=d$$, then $$SD=0$$ but for other values $$SD>0$$. Insufficient

Statement 2: Same as Statement 1, nothing mentioned about other variable. Insufficient

Combining 1 & 2: We know $$SD$$ is the variation from mean. So even if we assume that $$b$$, $$c$$, & $$d$$ are equal to mean (which will not be in this case) resulting in $$0$$ variation, $$a$$ will have variation.

So variation of $$a$$ from mean $$= 1-4=-3$$ and other variations $$= 0$$(assuming)

$$Variance =$$ average of square of variations $$= \frac{(-3)^2+0^2+0^2+0^2}{4}=2.25$$

So minimum $$SD= \sqrt{2.25}>1$$ (square root of any number greater than $$1$$ will be greater than $$1$$). Hence Sufficient

Option C
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Re: If the average (arithmetic mean) of a, b, c, and d is m, is their stan [#permalink]

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Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The average of $$a,b,c$$ and $$d$$ is
$$\frac{( a + b + c + d )}{4} = m$$.
Since we have 5 variables and 1 equation, E is most likely to be the answer. So, we should consider conditions 1) and 2) together first.

Conditions 1) and 2):
Standard deviation
= $$\frac{{√(a–m)^2 + (b-m)^2 + (c-m)^2 + (d-m)^2}}{4}$$
≥ $$√ (\frac{3^2}{4} )$$
= $$√ (\frac{9}{4} )$$
= $$\frac{3}{2}$$
> $$1$$
Both conditions 1) and 2) together are sufficient.

Since this is a statistics question (one of the key question areas), we should also consider choices A and B by CMT 4(A).

Condition 1):
If $$a = b = c = d = 1$$, the standard deviation is $$0<1$$, and the answer is ‘no’.
If $$a = 1, b = 4, c = 4, d = 7$$, the standard deviation is $$4.5 > 1$$, and the answer is ‘yes’.
Since we do not have a unique answer, condition 1) is not sufficient.

Condition 2):
If $$a = b = c = d = 4, m=4$$, the standard deviation is $$0<1$$, and the answer is
‘no’.
If $$a = 1, b = 4, c = 4, d = 7, m=4$$, the standard deviation is $$4.5 > 1$$, and the answer is ‘yes’.
Since we do not have a unique answer, condition 2) is not sufficient.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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Re: If the average (arithmetic mean) of a, b, c, and d is m, is their stan   [#permalink] 03 Jan 2018, 00:38
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text:
# Solving Ebufgne!

Misc problem in ECTF’16 created by skarthik and me. ez 100 points!

The problem provides a pcapng file which contains transfer of a zip file via tcp. After considering the zip file and trying to extract it, we see that it gives a error saying that the start of central directory was not found.

Now the file is correctly recognized as zip and the signature/header error is not show implying both of them are valid.

File is organized as:

--26 bytes(header)--Data--22 bytes(signature)--

This implies that something is wrong with the middle data. When we go ahead to fix the given central directory header, we realize that the bytes are actually swapped in place with each other. This has to do a bit with the challenge description.

Ebufgne -> Befunge which is basically an esoteric language developed in 1993 which used the symbol backward solidus i.e \ as a command to swap the top 2 bytes on the stack. Even if this information is not available, we can still solve the problem looking at the directory header.

On realizing this, we only have to write a script to swap bytes. After that we get a password encrypted zip file. But the password was already transmitted in the pcap as 103%secure. We use that and we get the flag!

There are 2 files in the zip: fl4g and d4t4. fl4g contains the flag :)

Below is the solver code:

# Writing the file.
def scramble(f):
current_position = f.tell()
f.seek(0, 2)
file_size = f.tell()
f.seek(current_position)

other_data = ""

while (file_size-f.tell()>=25):
other_data += b
other_data += a

return (other_data, end_of_dir)

f = open("DaaS.zip")
(zip_data, end_sig) = scramble(f)

new_file = open("problem.zip","w")


text:
# Question 8 Suppose that you have $1000 to invest, and you can invest it in stocks or bonds. Each ... Question 8 Suppose that you have$1000 to invest, and you can invest it in stocks or bonds. Each month, bonds yield a certain return of 0.8%. Each month, stocks yield a risky return of 1.8% with probability 0.7 and-1.1% with probability 0.3. Assume returns are independent across months. You choose your portfolio as suggested by Benartzi & Thaler. Let x be the change in your portfolio's value between now and the next time you evaluate your portfolio. Of course x will be risky _ that is, your choice will generate a lottery over possible outcomes for x. For any lottery (xi,pi;...,xN,pN), you evaluate this lottery according to prospective utility where v(xi) 2.25% ifXi 0. (a) Suppose you plan to evaluate your portfolio after one month. If you invest in all bonds, what is the lottery over x? If you invest in all stocks, what is the lottery over x? Which do you prefer, all bonds or all stocks?

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text:
# Could kernel density map be recovered with known kernel function?

I'm confused whether a KDE map could be recovered with known kernel function. The KDE map (with no noise) generated with fast fourier trasnformation (FFT) could be recovered on very high accuracy (<10e-3) in my experiment. The equations are as follows:

G(u,v)=H(u,v)*F(u,v)

F(u,v)=G(u,v)/H(u,v)

G(u,v) is the fourier transformation of density map

F(u,v) is the fourier transformation of original map

H(u,v) is the fourier transformation of kernel density function

But the KDE map generated by regular convolution method is very different to be recovered. I'm confused about those opposite answers. Anybody could tell "whether or not a kernel density map could be recovered?"

Many thanks!


text:
# Cotangent

If the angle α is acute, and cotg α = 1/3. Determine the value of sin α, cos α, tg α.

Result

sin α =  0.95
cos α =  0.32
tg α =  3

#### Solution:

$cotg \alpha = \dfrac { \cos \alpha }{ \sin \alpha } \ \\ \ \\ \sin \alpha = \dfrac{ 3}{ \sqrt{ 1^2 + 3^2}} = 0.95$
$\cos \alpha = \dfrac{ 1 }{ \sqrt{ 1^2 + 3^2}} = 0.32$
$tg \ \alpha = \dfrac{1}{ cotg \alpha} = \dfrac{ 3}{ 1} = 3$

Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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Tips to related online calculators
Most natural application of trigonometry and trigonometric functions is a calculation of the triangles. Common and less common calculations of different types of triangles offers our triangle calculator. Word trigonometry comes from Greek and literally means triangle calculation.

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Is true equality? ?
13. Theorem prove
We want to prove the sentence: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?
14. Is complex
Are these numbers 2i, 4i, 2i + 1, 8i, 2i + 3, 4 + 7i, 8i, 8i + 4, 5i, 6i, 3i complex?
15. Calculation
How much is sum of square root of six and the square root of 225?
16. Sequence 3
Write the first 5 members of an arithmetic sequence: a4=-35, a11=-105.
17. Linsys2
Solve two equations with two unknowns: 400x+120y=147.2 350x+200y=144


text:
## Journal of Symbolic Logic

### The Largest Countable Inductive Set is a Mouse Set

Mitch Rudominer

#### Abstract

Let $\kappa^\mathbb{R}$ be the least ordinal $\kappa$ such that L$_\kappa(\mathbb{R})$ is admissible. Let $A = \{x \in \mathbb{R} \mid (\exists\alpha < \kappa^\mathbb{R})$ such that x is ordinal definable in $L_\alpha (\mathbb{R})\}$. It is well known that (assuming determinacy) A is the largest countable inductive set of reals. Let T be the theory: ZFC - Replacement + "There exists $\omega$ Woodin cardinals which are cofinal in the ordinals." T has consistency strength weaker than that of the theory ZFC + "There exists $\omega$ Woodin cardinals", but stronger than that of the theory ZFC + "There exists n Woodin Cardinals", for each $n \in \omega$. Let $\mathcal{M}$ be the canonical, minimal inner model for the theory T. In this paper we show that A = $\mathbb{R} \cap \mathcal{M}$. Since $\mathcal{M}$ is a mouse, we say that A is a mouse set. As an application, we use our characterization of A to give an inner-model-theoretic proof of a theorem of Martin which states that for all n, every $\Sigma^*_n$ real is in A.

#### Article information

Source
J. Symbolic Logic, Volume 64, Issue 2 (1999), 443-459.

Dates
First available in Project Euclid: 6 July 2007

Permanent link to this document
https://projecteuclid.org/euclid.jsl/1183745786

Mathematical Reviews number (MathSciNet)
MR1777763

Zentralblatt MATH identifier
0930.03059

JSTOR

#### Citation

Rudominer, Mitch. The Largest Countable Inductive Set is a Mouse Set. J. Symbolic Logic 64 (1999), no. 2, 443--459. https://projecteuclid.org/euclid.jsl/1183745786


text:
## Topic outline

• #### MICROBIOLOGY COURSE

This comprehensive course contains 26 sections with 153 self-grading lessons, including final assessments. The first section is available to try for free.

This Trial Site is for evaluation purposes and for trying out in your LMS only. User names will be visible to everyone using it, so don't use this trial site with real students or for actual assignments! To view each lesson, POP-UPs MUST BE ENABLED in your browser.

• #### An Invisible World - Final Assessment

This is the final assessment for An Invisible World. (Available when you subscribe.)

• This topic

#### Subscribe & Try the FULL COURSE FREE for 14 days!

The full course includes all 153, including 26 Final Assessments with Instructor Answer Keys.

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Once you subscribe, we'll create your private course. Once you receive your course you'll have 14 days to try the course free. You can cancel anytime during your free trial and you won't be charged.

• #### How We See the Invisible World

2.1. The Properties of Light
2.2. Peering Into the Invisible World
2.3. Instruments of Microscopy
2.4. Staining Microscopic Specimens
Final Assessment

• #### The Cell

3.1. Spontaneous Generation
3.2. Foundations of Modern Cell Theory
3.3. Unique Characteristics of Prokaryotic Cells
3.4. Unique Characteristics of Eukaryotic Cells
Final Assessment

• #### Prokaryotic Diversity

4.1. Prokaryote Habitats, Relationships, and Microbiomes
4.2. Proteobacteria
4.3. Nonproteobacteria Gram-Negative Bacteria and Phototrophic Bacteria
4.4. Gram-Positive Bacteria
4.5. Deeply Branching Bacteria
4.6. Archaea
Final Assessment

• #### The Eukaryotes of Microbiology

5.1. Unicellular Eukaryotic Parasites
5.2. Parasitic Helminths
5.3. Fungi
5.4. Algae
5.5. Lichens
Final Assessment

• #### Acellular Pathogens

6.1. Viruses
6.2. The Viral Life Cycle
6.3. Isolation, Culture, and Identification of Viruses
6.4. Viroids, Virusoids, and Prions
Final Assessment

• #### Microbial Biochemistry

7.1. Organic Molecules
7.2. Carbohydrates
7.3. Lipids
7.4. Proteins
7.5. Using Biochemistry to Identify Microorganisms
Final Assessment

• #### Microbial Metabolism

8.1. Energy, Matter, and Enzymes
8.2. Catabolism of Carbohydrates
8.3. Cellular Respiration
8.4. Fermentation
8.5. Catabolism of Lipids and Proteins
8.6. Photosynthesis
8.7. Biogeochemical Cycles
Final Assessment

• #### Microbial Growth

9.1. How Microbes Grow
9.2. Oxygen Requirements for Microbial Growth
9.3. The Effects of pH on Microbial Growth
9.4. Temperature and Microbial Growth
9.5. Other Environmental Conditions that Affect Growth
9.6. Media Used for Bacterial Growth
Final Assessment

• #### Biochemistry of the Genome

10.1. Using Microbiology to Discover the Secrets of Life
10.2. Structure and Function of DNA
10.3. Structure and Function of RNA
10.4. Structure and Function of Cellular Genomes
Final Assessment

• #### Mechanisms of Microbial Genetics

11.1. The Functions of Genetic Material
11.2. DNA Replication
11.3. RNA Transcription
11.4. Protein Synthesis (Translation)
11.5. Mutations
11.6. How Asexual Prokaryotes Achieve Genetic Diversity
11.7. Gene Regulation: Operon Theory
Final Assessment

• #### Modern Applications of Microbial Genetics

12.1. Microbes and the Tools of Genetic Engineering
12.2. Visualizing and Characterizing DNA, RNA, and Protein
12.3. Whole Genome Methods and Pharmaceutical Applications of Genetic Engineering
12.4. Gene Therapy
Final Assessment

• #### Control of Microbial Growth

13.1. Controlling Microbial Growth
13.2. Using Physical Methods to Control Microorganisms
13.3. Using Chemicals to Control Microorganisms
13.4. Testing the Effectiveness of Antiseptics and Disinfectants
Final Assessment

• #### Antimicrobial Drugs

14.1. History of Chemotherapy and Antimicrobial Discovery
14.2. Fundamentals of Antimicrobial Chemotherapy
14.3. Mechanisms of Antibacterial Drugs
14.4. Mechanisms of Other Antimicrobial Drugs
14.5. Drug Resistance
14.6. Testing the Effectiveness of Antimicrobials
14.7. Current Strategies for Antimicrobial Discovery
Final Assessment

• #### Microbial Mechanisms of Pathogenicity

15.1. Characteristics of Infectious Disease
15.2. How Pathogens Cause Disease
15.3. Virulence Factors of Bacterial and Viral Pathogens
15.4. Virulence Factors of Eukaryotic Pathogens
Final Assessment

• #### Disease and Epidemiology

16.1. The Language of Epidemiologists
16.2. Tracking Infectious Diseases
16.3. Modes of Disease Transmission
16.4. Global Public Health
Final Assessment

• #### Innate Nonspecific Host Defenses

17.1. Physical Defenses
17.2. Chemical Defenses
17.3. Cellular Defenses
17.4. Pathogen Recognition and Phagocytosis
17.5. Inflammation and Fever
Final Assessment

• #### Adaptive Specific Host Defenses

18.1. Overview of Specific Adaptive Immunity
18.2. Major Histocompatibility Complexes and Antigen-Presenting Cells
18.3. T Lymphocytes and Cellular Immunity
18.4. B Lymphocytes and Humoral Immunity
18.5. Vaccines
Final Assessment

• #### Diseases of the Immune System

19.1. Hypersensitivities
19.2. Autoimmune Disorders
19.3. Organ Transplantation and Rejection
19.4. Immunodeficiency
19.5. Cancer Immunobiology and Immunotherapy
Final Assessment

• #### Laboratory Analysis of the Immune Response

20.1. Polyclonal and Monoclonal Antibody Production
20.2. Detecting Antigen-Antibody Complexes
20.3. Agglutination Assays
20.4. EIAs and ELISAs
20.5. Fluorescent Antibody Techniques
Final Assessment

• #### Skin and Eye Infections

21.1. Anatomy and Normal Microbiota of the Skin and Eyes
21.2. Bacterial Infections of the Skin and Eyes
21.3. Viral Infections of the Skin and Eyes
21.4. Mycoses of the Skin
21.5. Protozoan and Helminthic Infections of the Skin and Eyes
Final Assessment

• #### Respiratory System Infections

22.1. Anatomy and Normal Microbiota of the Respiratory Tract
22.2. Bacterial Infections of the Respiratory Tract
22.3. Viral Infections of the Respiratory Tract
22.4. Respiratory Mycoses
Final Assessment

• #### Urogenital System Infections

23.1. Anatomy and Normal Microbiota of the Urogenital Tract
23.2. Bacterial Infections of the Urinary System
23.3. Bacterial Infections of the Reproductive System
23.4. Viral Infections of the Reproductive System
23.5. Fungal Infections of the Reproductive System
23.6. Protozoan Infections of the Urogenital System
Final Assessment

• #### Digestive System Infections

24.1. Anatomy and Normal Microbiota of the Digestive System
24.2. Microbial Diseases of the Mouth and Oral Cavity
24.3. Bacterial Infections of the Gastrointestinal Tract
24.4. Viral Infections of the Gastrointestinal Tract
24.5. Protozoan Infections of the Gastrointestinal Tract
24.6. Helminthic Infections of the Gastrointestinal Tract
Final Assessment

• #### Circulatory and Lymphatic System Infections

25.1. Anatomy of the Circulatory and Lymphatic Systems
25.2. Bacterial Infections of the Circulatory and Lymphatic Systems
25.3. Viral Infections of the Circulatory and Lymphatic Systems
25.4. Parasitic Infections of the Circulatory and Lymphatic Systems
Final Assessment

• #### Nervous System Infections

26.1. Anatomy of the Nervous System
26.2. Bacterial Diseases of the Nervous System
26.3. Acellular Diseases of the Nervous System
26.4. Fungal and Parasitic Diseases of the Nervous System
Final Assessment


text:
Localization of energy in FPU chains
by Luisa Berchialla, Luigi Galgani and Antonio Giorgilli

Abstract: We revisit the celebrated model of Fermi, Pasta and Ulam with the aim of investigating, by numerical computations, the trend towards equipartition in the thermodynamic limit. We concentrate our attention on a particular class of initial conditions, namely, with all the energy on the first mode or the first few modes. We observe that the approach to equipartition occurs on two different time scales: in a short time the energy spreads up by forming a packet involving all low--frequency modes up to a cutoff frequency $\omega_c$, while a much longer time is required in order to reach equipartition, if any. In this sense one has an energy localization with respect to frequency. The crucial point is that our numerical computations suggest that this phenomenon of a fast formation of a natural packet survives in the thermodynamic limit. More precisely we conjecture that the cutoff frequency $\omega_c$ is a function of the specific energy $\epsilon = E/N$, where $E$ and $N$ are the total energy and the number of particles, respectively. Equivalently, there should exist a function $\epsilon_c(\omega)$, representing the minimal specific energy at which the natural packet extends up to frequency $\omega$. The time required for the fast formation of the natural packet is also investigated.

preprint (2002)

G-zipped file;   Zip compressed file;


text:
Home > Pre-Algebra calculators > Cube root of a number using long division method calculator

 Solve any problem (step by step solutions) Input table (Matrix, Statistics)
Mode :
SolutionHelp
Solution
Problem: cube root long division 27 [ Calculator, Method and examples ]

Solution:
Your problem -> cube root long division 27

Solution is
 3 9 27 27 --- 0

Number = 27
Cube Root = 3

Step by step solution :
Step-1 :
Make pair of digits of given number starting with digit at one's place. Put bar on each pair.
 27

Step-2 :
Now leftmost digits is 27. Now find the largest number whose cube is <= 27

It is 3, whose cube is 27. Write 3 as quotient and subtract 27
 3 9 27 27 0

Solution provided by AtoZmath.com
Any wrong solution, solution improvement, feedback then Submit Here
Want to know about AtoZmath.com and me


text:
# Difference between revisions of "2018 AMC 10B Problems/Problem 15"

A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper? $[asy]defaultpen(fontsize(10pt)); filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); dot((-3,3)); label("A",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label('w',(-1,-1),SW); label('w',(1,-1),SE); draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle); draw((4.5,0)--(8.5,0)); draw((6.5,2)--(6.5,-2)); label("A",(6.5,0),NW); dot((6.5,0)); [/asy]$

$\textbf{(A) } 2(w+h)^2 \qquad \textbf{(B) } \frac{(w+h)^2}2 \qquad \textbf{(C) } 2w^2+4wh \qquad \textbf{(D) } 2w^2 \qquad \textbf{(E) } w^2h$

## Solution

Consider one-quarter of the image (the wrapping paper is divided up into 4 congruent squares). The length of each dotted line is $h$. The area of the rectangle that is $w$ by $h$ is $wh$. The combined figure of the two triangles with base $h$ is a square with $h$ as its diagonal. Using the Pythagorean Theorem, each side of this square is $\sqrt{\frac{h^2}{2}}$. Thus, the area is the side length squared which is $\frac{h^2}{2}$. Similarly, the combined figure of the two triangles with base $w$ is a square with area $\frac{w^2}{2}$. Adding all of these together, we get $\frac{w^2}{2} + \frac{h^2}{2} + wh$. Since we have four of these areas in the entire wrapping paper, we multiply this by 4, getting $4(\frac{w^2}{2} + \frac{h^2}{2} + wh) = 2(w^2 + h^2 + 2wh) = \textbf{(A) } 2(w+h)^2 \qquad$. Solution by HydroQuantum


text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\text{Set Builder Notation: } \{x|x\le2\} \\\text{Interval Notation: } (-\infty,2]$
The given graph shows all the numbers that are less than or equal to $2.$ Hence, the given graph of inequality can be described in the following formats: \begin{array}{l}\require{cancel} \text{Set Builder Notation: } \{x|x\le2\} \\\text{Interval Notation: } (-\infty,2] .\end{array}


text:
?

Free Version
Difficult

Absolute Value From Graph

ACTMAT-FOEVBY

The graph on the number line below represents the solution to which of the following absolute value inequalities?

A

$|y-1|\ge 3$

B

$|y-1|\le 3$

C

$|y-3|\ge 1$

D

$|y+3|\le 1$

E

$|y+2|\le 4$


text:
Score bar

Your learning graph should look something like this by now. The score bar above shows how many of the possible points you have achieved so far. The points from the challenge tasks are displayed separately from those from the main and support tasks, since support tasks shall only compensate for the lost points from main tasks. Tasks of the type "information station" like this one do not provide any points since no solution is entered.

# Unassigned

Learning
5

Fill in the Blanks: Angle theorems

The image shows two parallel straight lines intersected by a third line. Fill in the blanks below in order to determine the size of angle $\beta$.

# Angles

Training
7

Vector (exact value): Angles

The image shows two pairs of parallel lines intersected by a fifth line. The size of angle $\alpha$ is 45°, $\beta$ is 65°. Determine the size (as numbers, without °) of the two purple angles and complete the checkboxes below.

# Intercept theorem

Training
7

Multiple Choice: Circle

Below, select all true statements about the circle in the image. The unit of length is centimeters.

# Circle

Training
7

Vector (Interval): Model the surface of the lake

The image shows lake Bracciano in Italy, which has an almost circular surface. Calculate the approximate area of the lakes surface. Fill out the answer fields below. All length units are in kilometers, the surface is in square kilometers. Round each to one decimal number. This task type is handy for tasks that could encounter rounding differences or whenever the user has to model or measure some quantity or size.

# Circle

Modeling
7

Given the quadratic function $f(x)= x^2-4x-5$. Determine all its zeros. The task type "set" does not take into account the order in which the answers are given (just like a set is commutative). Insert all results in the box below.


text:
# 3.3. Submodels

The genetic sub-model introduces a source of intra-specific variability of quantitative traits in OSMOSE, through additive genetic variance $$\sigma^2_{A,Z}$$ and expression variance $$\sigma^2_{e,Z}$$ and parental gene inheritance. The individuals are considered diploid hermaphrodites, i.e. without differentiation between males and females, the model being based on female life history. The genotypic values of the four heritable traits—maximum mass-specific ingestion rate $$I_{max}$$, gonado-somatic index $$r$$ , intercept and slope of linear maturation reaction ($$m_0$$ and $$m_1$$, respectively) norm—result from the expression of the corresponding functional loci. Neutral loci have no effect on individuals’ phenotype: their evolution is the result of random drift. Hereafter, the genetic sub-model is described for any of the four evolving traits, generically denoted $$Z$$.


text:
Download the gene or exon level RangedSummarizedExperiment-class objects provided by the recount project. Alternatively download the counts, metadata or file information for a given SRA study id. You can also download the sample bigWig files or the mean coverage bigWig file.

download_study(
project,
type = "rse-gene",
outdir = project,
version = 2,
...
)

## Arguments

project A character vector with one SRA study id. Specifies which files to download. The options are: rse-genethe gene-level RangedSummarizedExperiment-class object in a file named rse_gene.Rdata. rse-exonthe exon-level RangedSummarizedExperiment-class object in a file named rse_exon.Rdata. rse-jxthe exon-exon junction level RangedSummarizedExperiment-class object in a file named rse_jx.Rdata. rse-txthe transcript level RangedSummarizedExperiment-class object in a file named rse_tx.RData. counts-genethe gene-level counts in a tsv file named counts_gene.tsv.gz. counts-exonthe exon-level counts in a tsv file named counts_exon.tsv.gz. counts-jxthe exon-exon junction level counts in a tsv file named counts_jx.tsv.gz. phenotypethe phenotype data for the study in a tsv file named project.tsv. files-infothe files information for the given study (including md5sum hashes) in a tsv file named files_info.tsv. samplesone bigWig file per sample in the study. meanone mean bigWig file for the samples in the study, with each sample normalized to a 40 million 100 bp library using the total coverage sum (area under the coverage curve, AUC) for the given sample. allDownloads all the above types. Note that it might take some time if the project has many samples. When using type = 'all' a small delay will be added before each download request to avoid request issues. rse-fcDownloads the FANTOM-CAT/recount2 rse file described in Imada, Sanchez, et al., bioRxiv, 2019. The destination directory for the downloaded file(s). Alternatively check the SciServer section on the vignette to see how to access all the recount data via a R Jupyter Notebook. Whether to download the files or just get the download urls. A single integer specifying which version of the files to download. Valid options are 1 and 2, as described in https://jhubiostatistics.shinyapps.io/recount/ under the documentation tab. Briefly, version 1 are counts based on reduced exons while version 2 are based on disjoint exons. This argument mostly just matters for the exon counts. Defaults to version 2 (disjoint exons). Use version = 1 for backward compatability with exon counts prior to version 1.5.3 of the package. Additional arguments passed to download.

## Details

Check http://stackoverflow.com/a/34383991 if you need to find the effective URLs. For example, http://duffel.rail.bio/recount/DRP000366/bw/mean_DRP000366.bw points to a link from SciServer.

Transcript quantifications are described in Fu et al, bioRxiv, 2018. https://www.biorxiv.org/content/10.1101/247346v2

FANTOM-CAT/recount2 quantifications are described in Imada, Sanchez, et al., bioRxiv, 2019. https://www.biorxiv.org/content/10.1101/659490v1

## Examples

## Find the URL to download the RangedSummarizedExperiment for the

## See the actual URL
url
#> [1] "http://duffel.rail.bio/recount/v2/ERP001942/rse_gene.Rdata"
if (FALSE) {
## Download the example data included in the package for study SRP009615


text:
BluetoothError: (111, 'Connection refused')

I'm trying to connect raspberry pi 4 to my ubuntu laptop. I setup a rfcomm server on my laptop, the code is given below:

import bluetooth

server_sock = bluetooth.BluetoothSocket(bluetooth.RFCOMM)
server_sock.bind(("", bluetooth.PORT_ANY))
server_sock.listen(1)

port = server_sock.getsockname()[1]

uuid = "94f39d29-7d6d-437d-973b-fba39e49d4ee"

service_classes=[uuid, bluetooth.SERIAL_PORT_CLASS],
profiles=[bluetooth.SERIAL_PORT_PROFILE],
# protocols=[bluetooth.OBEX_UUID]
)

print("Waiting for connection on RFCOMM channel", port)

client_sock, client_info = server_sock.accept()
print("Accepted connection from", client_info)

try:
while True:
data = client_sock.recv(1024)
if not data:
break
except OSError:
pass

print("Disconnected.")

client_sock.close()
server_sock.close()
print("All done.")

And I setup a rfcomm client on the pi:

import sys
import bluetooth

uuid = ""
service_name='SampleServer'
service_matches = bluetooth.find_service(name=service_name)

if len(service_matches) == 0:
print("couldn't find the FooBar service")
sys.exit(0)

first_match = service_matches[0]
port = first_match["port"]
name = first_match["name"]
host = first_match["host"]

print("connecting to {} on {}".format(name, host))

sock=bluetooth.BluetoothSocket( bluetooth.RFCOMM )
sock.connect((host, port))
sock.send(b"hello!!")
sock.close()

I started the server first, however, when I ran the client, I found that the pi connected to my laptop for only a short time, and then the connection broke, and got the error: bluetooth.btcommon.BluetoothError: (111, 'Connection refused').

I've read some similar questions here, but they all seem to have set the mac address for server mistakenly, which is not what I did. So what's my problem here?

• "Connection refused" has three possible reasons. 1. Wrong address, 2 Right address, wrong port. 3. Right address, right port listener not listening, – Dougie May 28 '20 at 9:21
• Do you ask how to setup a rfcomm server on your laptop? – Ingo May 28 '20 at 17:11
• @Ingo not yet... Do you mean I may get a hint from that? – null May 29 '20 at 2:31
• I'm asking because I can't see a problem on the RasPi from your description. Configuring a rfcomm server is out of scope on Raspberry Pi. Can you please elaborate where the problem could be on the RasPi? – Ingo May 29 '20 at 18:08