8 = 2³

(2³)^x = 2^3x

2^3x = 2^x+6  <- when the bases are the same, you can set the exponents equal to each other and solve for x

3x=x+6

2x=6

x=3

First rewrite the two expressions so that they have the same base, and then compare their exponents.

$\displaystyle 27 = 3^{3}$   

$\displaystyle 27^2 = (3^{3})^2$

Combine exponents by multiplying: $\displaystyle (3^{3})^2 = 3^6$

This is the same as the first given expression, so the two expressions are equal.

$\displaystyle 32$ can be written as $\displaystyle 2^5.$ 

Since there is a common base of $\displaystyle 2$, we can say

 $\displaystyle 2^x=2^5$ or $\displaystyle x=5$. 

The basees don't match.

However: 

$\displaystyle \frac{1}{3}=3^{-1}$ thus we can rewrite the expression as $\displaystyle \frac{1}{9}=\left(\frac{1}{3}\right)^2$.

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

So, $\displaystyle (3^{-1})^{2}$$\displaystyle =3^{-2}$. $\displaystyle x=-2$. 

The bases don't match.

However: 

$\displaystyle 4=\frac{1}{4}^{-1}$ and we recognize that $\displaystyle 256=4^4$.

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

$\displaystyle \left(\frac{1}{4}^{-1}\right)^4=\frac{1}{4}^{-4}$.  

 $\displaystyle x=-4$

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 $\displaystyle x=\frac{1}{10}$.