# How do you solve a^x = 10^(2x+1)?

Apr 5, 2016

$x = \frac{- \log \left(10\right)}{2 \log \left(10\right) - \log \left(a\right)}$

#### Explanation:

$1$. Assuming you are trying to solve for $x$, start by taking the logarithm of both sides.

${a}^{x} = {10}^{2 x + 1}$

$\log \left({a}^{x}\right) = \log \left({10}^{2 x + 1}\right)$

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

$x \log \left(a\right) = \left(2 x + 1\right) \log \left(10\right)$

$3$. Expand the brackets.

$x \log \left(a\right) = 2 x \log \left(10\right) + \log \left(10\right)$

$4$. Move all terms with $x$ to one side of the equation with the terms with no $x$ to the other side.

$2 x \log \left(10\right) - x \log \left(a\right) = - \log \left(10\right)$

$5$. Factor out $x$.

$x \left(2 \log \left(10\right) - \log \left(a\right)\right) = - \log \left(10\right)$

$6$. Isolate for $x$.

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} x = \frac{- \log \left(10\right)}{2 \log \left(10\right) - \log \left(a\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$