When adding exponents, you want to factor out to make solving the question easier.

$\displaystyle 2^8+2^8$ we can factor out $\displaystyle 2^8$ to get 

$\displaystyle 2^8(1+1)=2^8(2)$.

We have the same base so we just apply the exponent rule for multiplication to get 

$\displaystyle 2^8(2)=2^{8+1}=2^9$.

Although each base is different, we can convert them to a common base of $\displaystyle 4.$ 

We know 

$\displaystyle 16^6=(4^2)^6=4^{12}$, 

$\displaystyle 64^4=(4^3)^4=4^{12}$,

and 

$\displaystyle 256^3=(4^4)^3=4^{12}$.

Remember to apply the power rule of exponents.

Therefore we have 

$\displaystyle 4^{12}+4^{12}+4^{12}+4^{12}$.

We can factor out $\displaystyle 4^{12}$ to get 

$\displaystyle 4^{12}(1+1+1+1)=4^{12}(4)=4^{13}$.

When adding exponents, you want to factor out to make solving the question easier.

$\displaystyle 3^{2x}+3^{2x}+3^{2x}$ 

We can factor out $\displaystyle 3^{2x}$ to get 

$\displaystyle 3^{2x}(1+1+1)=3^{2x}*3$.

Now we can add exponents and therefore our answer is 

$\displaystyle 3^{2x+1}$.

Express $\displaystyle 9$ as a power of $\displaystyle 3$; that is: $\displaystyle 9=3^2$.

Then $\displaystyle 3^2 \cdot 3^n = 3^9$.

Using the properties of exponents, $\displaystyle 3^{n+2}=3^9$.

Therefore, $\displaystyle n+2=9$, so $\displaystyle n=7$.

Since we have two $\displaystyle x$’s in $\displaystyle 9^{(x + 5)} + 3^{2(x + 5)}$ we will need to combine the two terms.

For $\displaystyle 3^{2(x + 5) }$ this can be rewritten as

$\displaystyle (3^2)^{ (x + 5)} = 9^ {(x + 5)}$

So we have $\displaystyle 9^{ (x + 5) }+ 9^{ (x + 5)} = 162$.

Or $\displaystyle 2 (9^{ (x + 5)}) = 162$

Divide this by $\displaystyle 2$: $\displaystyle 9^{ (x + 5)} = 81 = 9^ 2$

Thus $\displaystyle x +5 = 2$ or $\displaystyle x = -3$

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

m and n are both less than zero and thus negative integers, giving us m² and n² as perfect squares. The only perfect squares with a difference of 7 is 16 – 9, therefore m = –4 and n = –3.

To simplify, we can rewrite the numerator using a common exponential base.

$\displaystyle \frac{7^5+7^4}{8}=\frac{7(7^4)+7^4}{8}$

Now, we can factor out the numerator.

$\displaystyle \frac{7(7^4)+7^4}{8}=\frac{7^4(7+1)}{8}=\frac{7^4(8)}{8}$

The eights cancel to give us our final answer.

$\displaystyle \frac{7^4(8)}{8}=7^4$

The correct answer can be found by subtracting exponents that have the same base. Whenever exponents with the same base are divided, you can subtract the exponent of the denominator from the exponent of the numerator as shown below to obtain the final answer:

$\displaystyle \frac{x^{7}y^{3}}{x^{3}} = x^{7-3}y^{3} = x^{4}y^{3}$

You do not do anything with the y exponent because it has no identical bases.

$\displaystyle 4^{8}=\left ( 2^{2}\right )^{8}=2^{16}$ 

Subtract the denominator exponent from the numerator's exponent, since they have the same base.

$\displaystyle \frac{2^{16}}{2^{10}}=2^{16-10}=2^{6}=64$

Begin by noting that the group (4²³ - 4²¹) has a common factor, namely 4²¹.  You can treat this like any other constant or variable and factor it out.  That would give you: 4²¹(4² - 1). Therefore, we know that:

3² * (4²³ - 4²¹) = 3² * 4²¹(4² - 1)

Now, 4² - 1 = 16 - 1 = 15 = 5 * 3.  Replace that in the original:

3² * 4²¹(4² - 1) = 3² * 4²¹(3 * 5)

Combining multiples withe same base, you get:

3³ * 4²¹ * 5