# How do you solve 2^(8x-3) = 5^(6x-9) ?

Feb 16, 2017

$x = \frac{3 - 9 {\log}_{2} 5}{8 - 6 {\log}_{2} 5} = \frac{3 {\log}_{5} 2 - 9}{8 {\log}_{5} 2 - 6}$.

#### Explanation:

For $a > 0$, if ${a}^{b} = {a}^{c}$ then $b = c$.

$5 = {2}^{{\log}_{2} 5}$

Thus

$\textcolor{w h i t e}{\implies} {2}^{8 x - 3} = {5}^{6 x - 9}$
$\implies {2}^{8 x - 3} = {\left({2}^{{\log}_{2} 5}\right)}^{6 x - 9}$
$\implies {2}^{8 x - 3} = {2}^{\left({\log}_{2} 5\right) \left(6 x - 9\right)}$

$\implies 8 x - 3 = \left({\log}_{2} 5\right) \left(6 x - 9\right)$
$\implies 8 x - 3 = 6 x {\log}_{2} 5 - 9 {\log}_{2} 5$

$\implies 8 x - 6 x {\log}_{2} 5 = 3 - 9 {\log}_{2} 5$
$\implies x \left(8 - 6 {\log}_{2} 5\right) = 3 - 9 {\log}_{2} 5$

$\implies x = \frac{3 - 9 {\log}_{2} 5}{8 - 6 {\log}_{2} 5}$

## Another way:

$2 = {5}^{{\log}_{5} 2}$

So

$\textcolor{w h i t e}{\implies} {2}^{8 x - 3} = {5}^{6 x - 9}$

$\implies {5}^{\left({\log}_{5} 2\right) \left(8 x - 3\right)} = {5}^{6 x - 9}$

$\implies 8 x {\log}_{5} 2 - 3 {\log}_{5} 2 = 6 x - 9$

$\implies x \left(8 {\log}_{5} 2 - 6\right) = 3 {\log}_{5} 2 - 9$

$\implies x = \frac{3 {\log}_{5} 2 - 9}{8 {\log}_{5} 2 - 6}$