# How do you solve Log_(1/4) (x) + log _2 (x^2) =3?

May 29, 2016

We must first put in a common base; this can be done by using the change of base rule ${\log}_{a} \left(n\right) = \log \frac{n}{\log} a$

#### Explanation:

$\frac{\log x}{\log} \left(\frac{1}{4}\right) + \log {x}^{2} / \log 2 = 3$

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

$\log \frac{x}{- 2 \log 2} + \log {x}^{2} / \log 2 = 3$

$\log \frac{x}{- 2 \log 2} + \frac{- 2 \log {x}^{2}}{- 2 \log 2} = 3$

${\log}_{\frac{1}{4}} \left(x\right) + {\log}_{\frac{1}{4}} \left({x}^{-} 4\right) = 3$

${\log}_{\frac{1}{4}} \left(x \times {x}^{-} 4\right) = 3$

${x}^{-} 3 = {\left(\frac{1}{4}\right)}^{3}$

$\frac{1}{x} ^ 3 = \frac{1}{64}$

${x}^{3} = 64$

$x = 4$

Hopefully this helps!