# What is x if 3ln2+ln(x^2)+2=4?

##### 2 Answers
Nov 8, 2015

$x = {e}^{1 - \frac{3}{2} \ln \left(2\right)}$

#### Explanation:

Isolate the term involving $x$:

$\ln \left({x}^{2}\right) = 4 - 2 - 3 \ln \left(2\right) = 2 - 3 \ln \left(2\right)$

Use the property of logarithm $\ln \left({a}^{b}\right) = b \ln \left(a\right)$:

$2 \ln \left(x\right) = 2 - 3 \ln \left(2\right)$

Isolate the term involving $x$ again:

$\ln \left(x\right) = 1 - \frac{3}{2} \ln \left(2\right)$

Take the exponential of both terms:

${e}^{\ln \left(x\right)} = {e}^{1 - \frac{3}{2} \ln \left(2\right)}$

Consider the fact that exponential and logarithm are inverse functions, and thus ${e}^{\ln \left(x\right)} = x$

$x = {e}^{1 - \frac{3}{2} \ln \left(2\right)}$

Nov 8, 2015

$x = \pm \frac{e \sqrt{2}}{4}$

#### Explanation:

$\left[1\right] \text{ } 3 \ln 2 + \ln \left({x}^{2}\right) + 2 = 4$

Subtract $2$ from both sides.

$\left[2\right] \text{ } 3 \ln 2 + \ln \left({x}^{2}\right) + 2 - 2 = 4 - 2$

$\left[3\right] \text{ } 3 \ln 2 + \ln \left({x}^{2}\right) = 2$

Property: $a {\log}_{b} m = {\log}_{b} {m}^{a}$

$\left[4\right] \text{ } \ln {2}^{3} + \ln \left({x}^{2}\right) = 2$

$\left[5\right] \text{ } \ln 8 + \ln \left({x}^{2}\right) = 2$

Property: ${\log}_{b} m + {\log}_{b} n = {\log}_{b} \left(m n\right)$

$\left[6\right] \text{ } \ln \left(8 {x}^{2}\right) = 2$

$\left[7\right] \text{ } {\log}_{e} \left(8 {x}^{2}\right) = 2$

Convert to exponential form.

$\left[8\right] \text{ } \Leftrightarrow {e}^{2} = 8 {x}^{2}$

Divide both sides by $8$.

$\left[9\right] \text{ } {e}^{2} / 8 = {x}^{2}$

Subtract ${e}^{2} / 8$ from both sides.

$\left[10\right] \text{ } {x}^{2} - {e}^{2} / 8 = 0$

Difference of two squares.

$\left[11\right] \text{ } \left(x + \sqrt{{e}^{2} / 8}\right) \left(x - \sqrt{{e}^{2} / 8}\right) = 0$

$\left[12\right] \text{ } \left(x + \frac{e}{2 \sqrt{2}}\right) \left(x - \frac{e}{2 \sqrt{2}}\right) = 0$

Rationalize.

$\left[13\right] \text{ } \left(x + \frac{e \sqrt{2}}{4}\right) \left(x - \frac{e \sqrt{2}}{4}\right) = 0$

Therefore: $\textcolor{b l u e}{x = \pm \frac{e \sqrt{2}}{4}}$