# How do you solve 4^(x+1)= 4^x+6?

Apr 5, 2016

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

#### Explanation:

Using the rule:

${a}^{b + c} = {a}^{b} \left({a}^{c}\right)$

We can rewrite ${4}^{x + 1}$ as

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

This makes the equation

$\implies 4 \left({4}^{x}\right) = {4}^{x} + 6$

Subtract ${4}^{x}$ from both sides.

$\implies 4 \left({4}^{x}\right) - {4}^{x} = 6$

Note that although this looks different than you may be used to, $4 \left({4}^{x}\right) - {4}^{x} = 3 \left({4}^{x}\right)$ for the same reason that $4 u - u = 3 u$.

$\implies 3 \left({4}^{x}\right) = 6$

Divide both sides by $3$.

$\implies {4}^{x} = 2$

Write $4$ as ${2}^{2}$.

$\implies {\left({2}^{2}\right)}^{x} = 2$

Simplify the left hand side using the rule:

${\left({a}^{b}\right)}^{c} = {a}^{b c}$

This yields

$\implies {2}^{2 x} = 2$

We now have two exponential terms with the same base. Thus, since they're equal, we know their exponents must also be equal.

$\implies {2}^{2 x} = {2}^{1}$

$\implies 2 x = 1$

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