# How do you solve 5^(x+3)>10^(x-6)?

Nov 8, 2016

$x < 6 + 9 {\log}_{2} 5$

#### Explanation:

${10}^{x - 6} = {2}^{x - 6} {5}^{x - 6}$ then

${5}^{x + 3} / \left({5}^{x - 6}\right) > {2}^{x - 6}$ but

${5}^{x + 3} / \left({5}^{x - 6}\right) = {5}^{x + 3 - \left(x - 6\right)} = {5}^{9}$ then

${5}^{9} > {2}^{x - 6}$ or

${5}^{9} {2}^{6} > {2}^{x}$ now aplying $\log$ to both sides the inequality is preserved because $\log$ is monotonic strict increasing. So

$9 {\log}_{2} 5 + 6 {\log}_{2} 2 > x$ or finally

$x < 6 + 9 {\log}_{2} 5$