Recall that \(\displaystyle 128=2^7\). 

With same base, we can write this equation: 

\(\displaystyle x+1=7\). 

By subtracting \(\displaystyle 1\) on both sides, \(\displaystyle x=6\). 

Since \(\displaystyle 32=2^5\) we can rewrite the expression.

With same base, let's set up an equation of \(\displaystyle x^2+4=5\).

By subtracting \(\displaystyle 4\) on both sides, we get \(\displaystyle x^2=1\).

Take the square root of both sides we get BOTH \(\displaystyle 1\) and \(\displaystyle -1\). 

They don't have the same base, however: \(\displaystyle 25=5^2\).

Then \(\displaystyle 25^4=(5^2)^4\). You would multiply the \(\displaystyle 2\) and the \(\displaystyle 4\) instead of adding.

\(\displaystyle 2\cdot 4=8\). 

\(\displaystyle x=8\)

There are two ways to go about this.

Method \(\displaystyle 1:\)

They don't have the same bases however: \(\displaystyle 16=4^2\). Then \(\displaystyle 16^6=(4^2)^6\)

You would multiply the \(\displaystyle 2\) and the \(\displaystyle 6\) instead of adding. We have \(\displaystyle 2\cdot 6=2x\)

Divide \(\displaystyle 2\) on both sides to get \(\displaystyle x=6\).

Method \(\displaystyle 2\):

We can change the base from \(\displaystyle 4\) to \(\displaystyle 16.\)

\(\displaystyle 4^{2x}=(4^2)^x=16^x\) 

This is the basic property of the product of power exponents. 

We have the same base so basically \(\displaystyle x=6\). 

Since we can write \(\displaystyle 1024^x=(2^{10})^x\). 

With same base we can set up an equation of \(\displaystyle 10x=1\) 

Divide both sides by \(\displaystyle 10\) and we get \(\displaystyle x=\frac{1}{10}\). 

\(\displaystyle 1024^x=(2^{10})^x\) 

We still don't have the same base however:  \(\displaystyle 2=\frac{1}{2}^{-1}\)

Then,

\(\displaystyle 1024^x=(2^{10})^x=\left[\left(\frac{1}{2}\right)^{-1}\right]^{10x}\).

With same base we can set up an equation of \(\displaystyle -10x=1\). 

Divide both sides by \(\displaystyle -10\) and we get \(\displaystyle x=\frac{-1}{10}\). 

In order to determine the relationship between the quantities, solve each quantity.

4³ is 4 * 4 * 4 = 64

3⁴ is 3 * 3 * 3 * 3 = 81

Therefore, Quantity B is greater.

(–1) ¹³⁷= –1   

–1 < 0

(–1) ^{odd #} always equals –1.

(–1) ^{even #} always equals +1.

Anything raised to negative power means \(\displaystyle 1\) over the base raised to the postive exponent. 

\(\displaystyle 2^{-5}=\frac{1}{2^5}=\frac{1}{32}\)

Let's all convert the bases to \(\displaystyle 2\).

\(\displaystyle 4^{12}=[2^2]^{12}=2^{24}\)

\(\displaystyle 64^4=(2^6)^4=2^{24}\)

\(\displaystyle 16^8=[2^4]^8=2^{32}\)

\(\displaystyle \left(\frac{1}{2}\right)^{^{-24}}\) This one may be intimidating but \(\displaystyle 2=\frac{1}{2}^{-1}\).

Therefore, 

\(\displaystyle \left(\left[\frac{1}{2}\right]^{-24}\right)^{-1}=2^{24}\)

\(\displaystyle 2^{24}\)

\(\displaystyle 16^8\) is not like the answers so this is the correct answer.