# How do you solve \log _ { 5} ( x + 1) - \log _ { \frac { 1} { 5} } ( 2x + 5) < \log _ { 5} ( 2x ^ { 2} - x + 9)?

Jun 23, 2018

$x < \frac{1}{2}$

#### Explanation:

Writing

$\ln \left(x + 1\right) - \ln \frac{2 x + 5}{\ln} \left(\frac{1}{5}\right) < \ln \frac{2 {x}^{2} - x + 9}{\ln} \left(5\right)$

we have used that ${\log}_{a} b = \ln \frac{b}{\ln} \left(a\right)$
and $\ln \left(\frac{1}{5}\right) = \ln \left(1\right) - \ln \left(5\right) = - \ln \left(5\right)$
so we get

$\ln \left(\left(x + 1\right) \left(2 x + 5\right)\right) < \ln \left(2 {x}^{2} - x + 9\right)$ since $\ln \left(5\right) > 0$

so we get

$2 {x}^{2} + 2 x + 5 x + 5 < 2 {x}^{2} - x + 9$

simplifying

$8 x < 4$

$x < \frac{1}{2}$