# How do you solve these logarithmic equations?

## $4 + {\log}_{9} \left(3 x - 7\right) = 6$ ${\log}_{2} \left(2 x\right) + {\log}_{2} x = 5$ $3 {\log}_{5} x - {\log}_{5} \left(5 x\right) = 3 - {\log}_{5} 25$

Nov 25, 2016

See below.

#### Explanation:

Before the logarithm application, the equations read

${9}^{4} \left(3 x - 7\right) = {9}^{6}$
$\left(2 x\right) x = {2}^{5}$
${x}^{3} / \left(5 x\right) = {5}^{3} / 25$

after that we have

$3 x - 7 = {9}^{2} \to x = \frac{{9}^{2} + 7}{3}$
${x}^{2} = {2}^{4} \to x = \pm {2}^{2}$
${x}^{2} = {5}^{2} \to x = \pm 5$

Nov 25, 2016

$4 + {\log}_{9} \left(3 x - 7\right) = 6$

${\log}_{9} \left(3 x - 7\right) = 2$

$3 x - 7 = 81$

$3 x = 89$

$x = \frac{89}{3}$

${\log}_{2} \left(2 x\right) + {\log}_{2} \left(x\right) = 5$

${\log}_{2} \left(2 {x}^{2}\right) = 5$

$2 {x}^{2} = {2}^{5} = 32$

${x}^{2} = 16$

$x = \pm 4$

$3 {\log}_{5} x - {\log}_{5} \left(5 x\right) = 3 - {\log}_{5} 25$

${\log}_{5} {x}^{3} - {\log}_{5} \left(5 x\right) = 3 - 2 = 1$

${\log}_{5} \left({x}^{3} / \left(5 x\right)\right) = {\log}_{5} \left({x}^{2} / 5\right) = 1$

${x}^{2} / 5 = 5$

${x}^{2} = 25$

$x = \pm 5$