# What is x if ln(3x^2) + ln(x^4) + ln(7) =0?

Mar 27, 2018

Use the logarithm laws.

$\ln \left(3 {x}^{2} \cdot {x}^{4} \cdot 7\right) = 0$

$21 {x}^{6} = {e}^{0}$

${x}^{6} = \frac{1}{21}$

$x = \pm \sqrt[6]{\frac{1}{21}}$

Hopefully this helps!

Mar 27, 2018

The solutions are $x = \pm \sqrt[6]{\frac{1}{21}}$.

(or $x = \pm {21}^{- \frac{1}{6}}$.)

#### Explanation:

Use this logarithm rule:

${\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{red}{x}\right) + {\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{b l u e}{y}\right) = {\log}_{\textcolor{g r e e n}{a}} \left(\textcolor{red}{x} \cdot \textcolor{b l u e}{y}\right)$

Here's this rule applied to our equation:

$\ln \left(\textcolor{red}{3 {x}^{2}}\right) + \ln \left(\textcolor{b l u e}{{x}^{4}}\right) + \ln \left(\textcolor{g r e e n}{7}\right) = 0$

$\ln \left(\textcolor{red}{3 {x}^{2}} \cdot \textcolor{b l u e}{{x}^{4}}\right) + \ln \left(\textcolor{g r e e n}{7}\right) = 0$

$\ln \left(\textcolor{red}{3} \textcolor{p u r p \le}{{x}^{6}}\right) + \ln \left(\textcolor{g r e e n}{7}\right) = 0$

$\ln \left(\textcolor{red}{3} \textcolor{p u r p \le}{{x}^{6}} \cdot \textcolor{g r e e n}{7}\right) = 0$

$\ln \left(\textcolor{b r o w n}{21} \textcolor{p u r p \le}{{x}^{6}}\right) = 0$

${\log}_{e} \left(\textcolor{b r o w n}{21} \textcolor{p u r p \le}{{x}^{6}}\right) = 0$

Convert to exponential form:

${e}^{0} = 21 {x}^{6}$

$1 = 21 {x}^{6}$

$\frac{1}{21} = {x}^{6}$

$\sqrt[6]{\frac{1}{21}} = x$

Since the root is an even power, we add a plus-or-minus sign:

$x = \pm \sqrt[6]{\frac{1}{21}}$

$x = \pm \sqrt[6]{{21}^{-} 1}$

$x = \pm {\left({21}^{-} 1\right)}^{\frac{1}{6}}$

$x = \pm {21}^{- \frac{1}{6}}$

You can check using a graphing calculator:

Since the values of the zeroes are the same as our answer, we are correct. Hope this helped!