# How do you solve log_5 2 + 2 log_5t= log_5(3-t)?

May 4, 2016

$t = - \frac{3}{2} \mathmr{and} t = 1$

#### Explanation:

Use Property : ${\log}_{b} \left(x y\right) = {\log}_{b} x + {\log}_{b} y$ and ${\log}_{b} {x}^{n} = n {\log}_{b} x$

${\log}_{5} 2 + 2 {\log}_{5} t = {\log}_{5} \left(3 - t\right)$

${\log}_{5} 2 + {\log}_{5} {t}^{2} = {\log}_{5} \left(3 - t\right)$

${\log}_{5} 2 {t}^{2} = {\log}_{5} \left(3 - t\right)$

$2 {t}^{2} = 3 - t$

$2 {t}^{2} + t - 3 = 0$

$\left(2 t + 3\right) \left(t - 1\right) = 0$

$t = - \frac{3}{2} \mathmr{and} t = 1$