# How do you solve 4^(2p)=4^(-2p-1)?

$p = - \frac{1}{4}$

#### Explanation:

Given equation

${4}^{2 p} = {4}^{- 2 p - 1}$

Comparing the powers on base $4$ we get

$2 p = - 2 p - 1$

$2 p + 2 p = - 1$

$4 p = - 1$

$p = - \frac{1}{4}$

Jul 28, 2018

$p = - \frac{1}{4}$

#### Explanation:

$\text{since both sides are expressed in base 4}$

$\text{we can equate the exponents}$

$2 p = - 2 p - 1$

$4 p = - 1 \Rightarrow p = - \frac{1}{4}$

Jul 28, 2018

$p = - \frac{1}{4}$

#### Explanation:

Since the bases are the same, we can equate the exponents. With this in mind, we now have:

$2 p = - 2 p - 1$

Next, let's add $2 p$ to both sides to get

$4 p = - 1$

Lastly, we can divide both sides by $4$ to get

$p = - \frac{1}{4}$

Hope this helps!