When adding exponents, we don't multiply the exponents but we try to factor to see if we simplify the addition problem. In this case, we can simplify it by factoring $\displaystyle 3^7$. We get $\displaystyle 3^7(1+1+1)=3^7(3)=3^8$.

When adding exponents, we don't multiply the exponents but we try to factor to see if we simplify the addition problem. In this case, we can simplify it by factoring $\displaystyle 5^7$. We get $\displaystyle 5^7(1+1+1+1+1)=5^7(5)=5^8$.

Although we have different bases, we do know $\displaystyle 4=2^2$. Therefore our expression is $\displaystyle 2^{12}+(2^2)^6=2^{12}+2^{12}$. Remember to apply the power rule of exponents. Then, now we can factor $\displaystyle 2^{12}$. $\displaystyle 2^{12}+2^{12}=2^{12}(1+1)=2^{12}(2)=2^{13}$

$\displaystyle \dpi{100} \small 25\left ( 5^{r} \right )$ is equal to $\displaystyle 5^{2}\left ( 5^{r}\right )$ which is equal to $\displaystyle \dpi{100} \small \left ( 5^{r+2} \right )$. If we compare this to the original equation we get $\displaystyle \dpi{100} \small r+2=s-2\rightarrow s=r+4$

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.