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\(\log_2(\log_3x)=2\log_4x\\ \log_2(\log_3x)=\frac{2\log_2x}{\log_24}\\ \log_2(\log_3x)=\log_2x\\ \log_3x=x\)
That has no real solutions. Has complex solutions but... i dont know them so. If u want them check this out:
https://www.wolframalpha.com/input?i=log_3%28x%29%3Dx

its ok to post questions that one thinks are fun and have a good solution even though i know the answer right? Its like sharing what I think is a good experience.

we have:

\(\log_2x=4\\ \boxed{2^4=x=16}\)

\( \log_{\frac{1}{2}} (16*5*1/10)=x\\ \log_{\frac{1}{2}} (8)=x\\ (\frac{1}{2})^x=8\\ \boxed{x=-3}\)

\(\angle ZNM + \angle ZMN + \angle PNM + \angle PMN + \angle NZM + \angle NPM= \) what we need to find. 
Lets rearrange what we need to find:
​​\((\angle ZNM + \angle ZMN + \angle NZM) + (\angle PNM + \angle PMN + \angle NPM)\) we can see they are angles of a triangle so:|
\(\boxed{180+180=360^\circ}\)

At least 5 means we need 1 in between 5 and 6. imagine this as a block. So including the block there are 4 total numbers. These numbers have \(4!=24\) ways to arrange them. Then the block has 2 permutations. There can be anyother number next to each other as it is always greater than 5 so:

\(24\times2=\boxed{48}\)

\(R=\frac{\rho l}{A}\) If \(l\) becomes \(l/2\) and everything else is the same then:
\(R_0=R/2\)
 

\(V=IR\\ R=\frac{V}{I}\\ R=\frac{500}{0.025}=20000\Omega\)
\(R_0=10000\Omega\)

So:

\(V_0=I_0R_0\\ I_0=\frac{V_0}{R_0}=\frac{175}{10000}\\ \boxed{I_0=17.5\text{ milliamps}}\)

Set the speed of the plane to k.

Set the distance to x.

A normal roundtrip is expressed as \(\frac{2x}{k}=\frac{10}{3}\), by the formula time = distance/speed.

A roundtrip with wind is \(\frac{x}{k+70}+\frac{x}{k-70}=\frac{23}{6}\), because the speed is 70 more when going to Jackson than when coming back.

To solve this system, \(\begin{cases}\frac{2x}{k}=\frac{10}{3} \\ \frac{x}{k+70}+\frac{x}{k-70}=\frac{23}{6} \end{cases}\). We want to solve for x, so we solve for k in terms of k. We should get \(k=\frac{3}{5}x\).

We plug this in, and solve. We get \(x=\frac{350\sqrt{69}}{9}\). So the distance is \(\frac{350\sqrt{69}}{9}\).

\(xy + x = \frac{3x + 2y + z + y + 2z}{3}\)

\(xy+x=\frac{3x+3y+3z}{3}\).

\(xy+x=x+y+z\)

\(xy=y+z\)

\(x=\frac{y+z}{y}\). Can also be written as: \(x=1+\frac{z}{y}\).

\(g(x)=f\left((x^2 - 2)/(x + 1)\right)\).

We can only input values from -8 to 8 in the function, so let's start from there.

\(-8\le\frac{{x}^{2}-2}{x+1}\le8\)

Consider there two seperately.

Case 1: \(-8\le\frac{{x}^{2}-2}{x+1}\)

\(\frac{{x}^{2}+8x+6}{x+1}\ge0\)

Use the method of intervals, or the wavy curve method. (I love that name ).

\({x}^{2}+8x+6\) has roots at \(-4 \pm \sqrt{10}\). 

so we get \(\frac{{x}^{2}+8x+6}{x+1}\ge0\) on the intervals \([-4-\sqrt{10}, -1)\cup[-4+\sqrt{10}, \infty)\). Remember, you can't plug -1 in so thats why it's not included.

Case 2: \(\frac{{x}^{2}-2}{x+1}\le8\). (I will spare some explanation here, its the same as above).

The interval turns out to be \((-\infty, 4-\sqrt{26}]\cup(1, 4+\sqrt{26}]\).

Now, we need to find the intersection of these intervals, since it satisfies both conditions.

We get \([-4-\sqrt{10}, 4-\sqrt{26}]\cup[-4+\sqrt{10}, 4+\sqrt{26}]\).

This interval has a width of 16. 

\(\frac{1}{64a^3 + 7} - 7 = 0\)

Add 7 to both sides: \(\frac{1}{64a^3 + 7} =7\)

Multiply both sides by 64a^3 + 7 to get rid of denominator: \(1 = 7(64a^3+7)\); \(1 = 448a^3+49\); \(-48 =448a^3\); \(-{3\over28}=a^3\)

Cube rooting both sides, you get \(a = ({-{3\over28}})^{1\over3}\), or noted as the cube root of -3/28.

\(\frac{a}{3} + 1 = \frac{a + 3}{a} + 1\)

Subtract 1 from both sides: \({a\over3}={a+3\over{a}}\)

Cross multiply: \(a^2=3a+9\); \(a^2-3a-9=0\)

Use quadratic formula: \(a = {3 \pm\sqrt{9+36}\over2}={3\pm3\sqrt{5}\over2}\)

t = cost of toast, b = cost of bagel, m = cost of muffin

2t + b = 3.15                              Equation 1

b + m = 3.5                                 Equation 2

t + 2b + 3m = 8.15                      Equation 3

Use elimination:

2*Equation 3 - Equation 1:

2t + 4b + 6m - 2t - b = 16.3 - 3.15; 3b + 6m = 13.15                      Equation 3.1

6*Equation2 - Equation 3.1:

6b + 6m - 3b - 6m = 21 - 13.15

3b = 7.85

b = 261.666666666... cents

So, a bagel costs (roughly) 262 cents.

Add the two equations together and you get 2a = 10, a = 5, which is your only solution.

p + r = 9 + p, subtract p from both sides, you get r = 9.

q - 2r = -2 + q, subtract q from both sides, you get -2r = -2, r = 1. 

1 is not equal to 9, so there may be a latex or copy and paste issue when transferring websites 

\(ab^4=48\), \(ab^2=4\)

\(ab^4=(ab^2)b^2=4b^2=48\)

\(b^2=12\), \(b=\pm2\sqrt{3}\)

If b has to be positive as specified possibly in your question, then type 2sqrt(3), if not, then type -2sqrt(3).

Interior angle sum of ABC is 180 degrees, so p + 2q + 6p - 5q + 14p + 8q = 180

Combine like terms: 21p + 5q = 180. 

21p = 180 - 5q

p = (180 - 5q)/21

(r, theta) in polar form, where theta indicates the direction and r, the length. Think of (2, pi/3) as a triangle with diagonal two and angle pi/3 = 60 degrees.

Thus, the base of the triangle, is 2/2 = 1, and then the height would be sqrt(3), so Point A has coordinates (1, sqrt(3)).

Point B likewise, in the fourth quadrant, would have the included angle be 30 degrees with a diagonal of 4. Thus, its coordinates would be (2sqrt(3), -2).