# How do you solve the exponential equation 2^(x+2)=16^x?

Feb 17, 2017

$x = \frac{2}{3}$

#### Explanation:

Note that $16 = {2}^{4}$

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

$\Rightarrow {\textcolor{red}{2}}^{x + 2} = {\textcolor{red}{2}}^{4 x}$

Since the bases are equal, that is $\textcolor{red}{2}$ then the exponents are equal.

$\text{solve } 4 x = x + 2$

$\Rightarrow 3 x = 2$

$\Rightarrow x = \frac{2}{3} \text{ is the solution}$

Feb 17, 2017

$x = \frac{2}{3}$

#### Explanation:

${2}^{x + 2} = {16}^{x}$

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

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

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

because the base on the LHS =2 and base on RHS=2,
so the exponents are equal

$x + 2 = 4 x$

$x - 4 x = - 2$

$- 3 x = - 2$

multiply both sides by $- 1$

$3 x = 2$

$x = \frac{2}{3}$

substitute$x = \frac{2}{3}$

${2}^{\left(\frac{2}{3}\right) + 2} = {16}^{\frac{2}{3}}$

${2}^{\frac{8}{3}} = {\left(2 \cdot 2 \cdot 2 \cdot 2\right)}^{\frac{2}{3}}$

${2}^{\frac{8}{3}} = {\left({2}^{4}\right)}^{\frac{2}{3}}$

${2}^{\frac{8}{3}} = {2}^{\frac{8}{3}}$