When an exponent is raised to the power of another exponent, just multiply the exponents together.

\(\displaystyle y = (x^a)^b = x^{ab}\)

\(\displaystyle y = [((-7x+12)^{1/3})^3]^7 = ((-7x+12)^{1/3})^{21}\) \(\displaystyle = (-7x+12)^7\)

First, you can begin to simplfy the numerator by converting all 3 expressions into base 2.

\(\displaystyle 2^\frac{5}{3}\cdot (2^2)^\frac{4}{3}\cdot (2^3)^\frac{1}{3}\), which simplifies to 

\(\displaystyle 2^3\cdot 2 =16\)

For the denominator, the same method applies. Convert the 25 into base 5, and when simplified becomes simply 5.

\(\displaystyle 5^{\frac{1}{3}}\cdot 5^{2\cdot \frac{1}{3}}\)

\(\displaystyle 5^{\frac{1}{3}+\frac{2}{3}}=5^1=5\)

The final simplified answer becomes:

 \(\displaystyle \frac{\text{Numerator}}{\text{Denominator}} = \frac{16}{5}\) 

Let's work through this equation involving exponents one term at a time. The first term we see is \(\displaystyle 8^\frac{2}{3}\), for which we can apply the following property:

\(\displaystyle a^\frac{c}{b}=\sqrt[b]{a^c}=(\sqrt[b]{a})^c\)

So if we plug our values into the formula for the property, we get:

\(\displaystyle 8^\frac{2}{3}=\sqrt[3]{8^2}=(\sqrt[3]{8})^2=(2)^2=4\)

Because \(\displaystyle 2^3=8\). Our next term is \(\displaystyle (2^2)^2\), for which we'll need the property:

\(\displaystyle (a^b)^c=a^{bc}\)

Using the values for our term, we have:

\(\displaystyle (2^2)^2=2^4=2\cdot 2\cdot 2\cdot 2=16\)

The third term of the equation is \(\displaystyle (3)^3\left(\frac{1}{3}\right)^3\), for which the quickest way to evaluate would be using the following property:

\(\displaystyle (a)^c(b)^c=(ab)^c\)

Using the values from our term, this gives us:

\(\displaystyle (3)^3(\frac{1}{3})^3=(3\cdot \frac{1}{3})^3=(1)^3=1\)

The next property we will need to consider for our fourth term is given below:

\(\displaystyle \frac{a^b}{a^c}=a^{b-c}\)

If we plug in the corresponding values from our term, we get:

\(\displaystyle \frac{5^{19}}{5^{17}}=5^{19-17}=5^2=25\)

Finally, our last term requires knowledge of the following simple property: Any number raised to the power of zero is 1. With this in mind, our last term becomes:

\(\displaystyle 647^0=1\)

Rewriting the equation with all of the values we've just evaluated, we obtain our final answer:

\(\displaystyle 4+16-1+25-1=43\)

To solve this problem, recall that you can set exponents equal to eachother if they have the same base.

See below:

\(\displaystyle 343=7^3\)

So, we have

\(\displaystyle 7^3=7^{15t-27}\)

Because both sides of this equation have a base of seven, we can set the exponents equal to eachother and solve for t.

\(\displaystyle 3=15t-27\)

\(\displaystyle 30=15t\)

\(\displaystyle t=2\)

We begin by taking the natural log of the equation:

\(\displaystyle \ln(2^{x/3}\cdot 3^{x/4}\cdot e^{-x})=\ln3\)

Simplifying the left side of the equation using the rules of logarithms gives:

\(\displaystyle \frac{x}{3}\ln2+\frac{x}{4}\ln3-x=\ln3\)

We group the x terms to get:

\(\displaystyle x\left(\frac{1}{3}\ln2+\frac{1}{4}\ln3-1\right)=\ln3\)

We reincorporate the exponents into the logarithms and use the identity property of the natural log to obtain:

\(\displaystyle x(\ln2^{1/3}+\ln3^{1/4}+\ln e^{-1})=\ln3\)

We combine the logarithms using the multiplication/sum rule to get:

\(\displaystyle x\ln(2^{1/3} 3^{1/4}e^{-1})=\ln3\)

We then solve for x:

\(\displaystyle x=\frac{\ln3}{\ln(2^{1/3}3^{1/4}e^{-1})}\approx-2.223\)

We begin by factoring out the term \(\displaystyle x^{1/2}\) to get:

\(\displaystyle x^{1/2}(x^2-12x+20)=0\)

This equation gives our first solution:

\(\displaystyle x^{1/2}=0\Rightarrow x=0\)

Then we check for more solutions:

\(\displaystyle \\x^2-12x+20=0 \\ \Rightarrow (x-10)(x-2)=0\\ \Rightarrow x=2,10\)

Therefore our solution is

\(\displaystyle x=0,2,10\)

Remember the denominator of a rational exponent is equivalent to the index of a root.

This should simplify quite nicely.

 \(\displaystyle \frac{3xz^{\frac{1}{3}}}{(6x)^{\frac{1}{2}}y^{2}z^{3}}=\frac{3x^{1/2}}{\sqrt{2\cdot3}y^2z^{8/3}}=\frac{\sqrt{3x}}{\sqrt{2}y^2\sqrt[3]{z^{8}}}\)

When \(\displaystyle x=2,y=3^{\frac{1}{4}}, z=8\) it gives us, 

\(\displaystyle \frac{\sqrt{3\cdot2}}{\sqrt{2}(3^{\frac{1}{4}})^2\sqrt[3]{8^{8}}}=\frac{\sqrt{3}}{(3^{\frac{2}{4}})2^{8}}=\frac{1}{2^8}=\frac{1}{256}\)

What does an exponent of one-third mean? Consider our expression and raise it to the third power. 

\(\displaystyle (125^\frac{1}{3})^3 = x^3\) 

Simplifying, we get: 

\(\displaystyle 125 = x^3\)

Thus, we are looking for a number that when cubed, we get \(\displaystyle 125\). Thus, we are discussing the cube root of \(\displaystyle 125\), or \(\displaystyle 5\). 

The derifintion of logarithm is:

\(\displaystyle log_{a}y=x \Leftrightarrow a^{x}=y.\)

In this problem,

\(\displaystyle a= 10, y=1000\) 

\(\displaystyle 10^{3}=1000\)

Therefore, \(\displaystyle x=3\)

Using the rules of logarithms, 

\(\displaystyle \log a - \log b = \log \bigg(\frac{a}{b}\bigg)\)

Hence,

 \(\displaystyle \log x^2 - \log 2x = log \bigg(\frac{x^2}{2x}\bigg) = log \bigg(\frac{x}{2}\bigg) = 2\)

So exponentiate both sides with a base 10:

\(\displaystyle 10^{log_{10} \frac{x}{2}} = 10^{2}\)

The exponent and the logarithm cancel out, leaving:  

\(\displaystyle \frac{x}{2}=10^2\)

\(\displaystyle x=100\cdot 2=200\)

This answer does not match any of the answer choices, therefore the answer is 'None of the other choices'.