# Question #072fe

Jan 23, 2017

#### Explanation:

I am going to assume base 10:

${\log}_{10} \left(\sqrt{10 {\left(x - 1\right)}^{2}}\right) - \frac{1}{3} {\log}_{10} \left({\left(x + 3\right)}^{2}\right) = \frac{1}{3}$

Remove the root from inside the first logarithm by multiplying by 1/3:

$\frac{1}{3} {\log}_{10} \left(10 {\left(x - 1\right)}^{2}\right) - \frac{1}{3} {\log}_{10} \left({\left(x + 3\right)}^{2}\right) = \frac{1}{3}$

Multiply both sides of the equation by 3:

${\log}_{10} \left(10 {\left(x - 1\right)}^{2}\right) - {\log}_{10} \left({\left(x + 3\right)}^{2}\right) = 1$

The difference of two logarithms is the same as division within the argument:

${\log}_{10} \left(10 {\left(x - 1\right)}^{2} / {\left(x + 3\right)}^{2}\right) = 1$

Eliminate the logarithm by making both sides the power of 10:

${10}^{{\log}_{10} \left(10 {\left(x - 1\right)}^{2} / {\left(x + 3\right)}^{2}\right)} = {10}^{1}$

On the left side the base and the logarithm are removed:

$10 {\left(x - 1\right)}^{2} / {\left(x + 3\right)}^{2} = {10}^{1}$

Divide both sides by 10:

${\left(x - 1\right)}^{2} / {\left(x + 3\right)}^{2} = 1$

Multiply both sides by ${\left(x + 3\right)}^{2}$:

${\left(x - 1\right)}^{2} = {\left(x + 3\right)}^{2}$

Expand the squares:

${x}^{2} - 2 x + 1 = {x}^{2} + 6 x + 9$

Add $- {x}^{2} - 6 x - 1$ to both sides:

$- 8 x = 8$

Divide by sides by -8:

$x = - 1$

Jan 23, 2017

$x = - \frac{5}{4}$

#### Explanation:

$\log \sqrt{10 {\left(x - 1\right)}^{2}} - \frac{1}{3} \log {\left(3 + x\right)}^{2} = \frac{1}{3}$

$\log \sqrt{10 {\left(x - 1\right)}^{2}} - \log {\left(3 + x\right)}^{\frac{2}{3}}$

$\log \left(\frac{\sqrt{10 {\left(x - 1\right)}^{2}}}{3 + x} ^ \left(\frac{2}{3}\right)\right) = \frac{1}{3}$

$\frac{\sqrt{10 {\left(x - 1\right)}^{2}}}{3 + x} ^ \left(\frac{2}{3}\right) = {10}^{\frac{1}{3}}$

$\sqrt{10 {\left(x - 1\right)}^{2}} = {10}^{\frac{1}{3}} \left({\left(3 + x\right)}^{\frac{2}{3}}\right)$

${\left(\sqrt{10 {\left(x - 1\right)}^{2}}\right)}^{3} = {\left({10}^{\frac{1}{3}} \left({\left(3 + x\right)}^{\frac{2}{3}}\right)\right)}^{3}$

$10 {\left(x - 1\right)}^{2} = 10 {\left(3 + x\right)}^{2}$

${\left(x - 1\right)}^{2} = {\left(3 + x\right)}^{2}$

${x}^{2} - 2 x - 1 = {x}^{2} + 6 x + 9$

$8 x + 10 = 0$

$8 x = - 10$

$x = - \frac{10}{8} = - \frac{5}{4}$