How do you solve 8^x = 4 times 12^(2x)?

Sep 16, 2016

$x = - \frac{2 \ln \left(2\right)}{\ln \left(18\right)}$

Explanation:

We have: ${8}^{x} = 4 \times {12}^{2 x}$

Let's apply the natural logarithm to both sides of the equation:

$\implies \ln \left({8}^{x}\right) = \ln \left(4 \times {12}^{2 x}\right)$

Using the laws of logarithms:

$\implies x \ln \left(8\right) = \ln \left(4\right) + 2 x \ln \left(12\right)$

Let's express some numbers in terms of $2$:

$\implies x \ln \left({2}^{3}\right) = \ln \left({2}^{2}\right) + 2 x \ln \left(12\right)$

$\implies 3 x \ln \left(2\right) = 2 \ln \left(2\right) + 2 x \ln \left(12\right)$

$\implies 3 x \ln \left(2\right) - 2 x \ln \left(12\right) = 2 \ln \left(2\right)$

$\implies x \left(3 \ln \left(2\right) - 2 \ln \left(12\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(\ln \left({2}^{3}\right) - \ln \left({12}^{2}\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(\ln \left(\frac{{2}^{3}}{{12}^{2}}\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(\ln \left(\frac{8}{144}\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(\ln \left(\frac{1}{18}\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(\ln \left(1\right) - \ln \left(18\right)\right) = 2 \ln \left(2\right)$

$\implies x \left(0 - \ln \left(18\right)\right) = 2 \ln \left(2\right)$

$\implies - x \ln \left(18\right) = 2 \ln \left(2\right)$

$\implies x \ln \left(18\right) = - 2 \ln \left(2\right)$

$\implies x = - \frac{2 \ln \left(2\right)}{\ln \left(18\right)}$

Therefore, the solution to the equation is $x = - \frac{2 \ln \left(2\right)}{\ln \left(18\right)}$.