# How do you solve the exponential equation 16^(x+1)=4^(4x+1)?

Mar 7, 2017

$x = \frac{1}{2}$

#### Explanation:

the key to these sorts of questions is to change the base numbers so that they are the same.

${16}^{x + 1} = {4}^{4 x + 1}$

now $16 = {4}^{2}$

so ${\left({4}^{2}\right)}^{x + 1} = {4}^{4 x + 1}$

${4}^{2 x + 2} = {4}^{4 x + 1}$

$\implies 2 x + 2 = 4 x + 1$

$\implies 2 - 1 = 4 x - 2 x$
$1 = 2 x$

$\therefore x = \frac{1}{2}$