# How do you solve  16^(x - 4) = 3^(3 - x)?

Mar 16, 2016

$x \approx 3.72$

#### Explanation:

$1$. Since the left and right sides of the equation do not have the same base, start by taking the log of both sides.

${16}^{x - 4} = {3}^{3 - x}$

$\log \left({16}^{x - 4}\right) = \log \left({3}^{3 - x}\right)$

$2$. Use the log property, ${\log}_{\textcolor{p u r p \le}{b}} \left({\textcolor{red}{m}}^{\textcolor{b l u e}{n}}\right) = \textcolor{b l u e}{n} \cdot {\log}_{\textcolor{p u r p \le}{b}} \left(\textcolor{red}{m}\right)$, to simplify both sides of the equation.

$\left(x - 4\right) \log 16 = \left(3 - x\right) \log 3$

$3$. Expand the brackets.

$x \log 16 - 4 \log 16 = 3 \log 3 - x \log 3$

$4$. Group all like terms together such that the terms with the variable, $x$, are on the left and the ones without on the right.

$x \log 16 + x \log 3 = 3 \log 3 + 4 \log 16$

$5$. Factor out $x$ from the terms on the left side of the equation.

$x \left(\log 16 + \log 3\right) = 3 \log 3 + 4 \log 16$

$6$. Solve for $x$.

$x = \frac{3 \log 3 + 4 \log 16}{\log 16 + \log 3}$

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x \approx 3.72 \textcolor{w h i t e}{\frac{a}{a}} |}}}$