# How do you solve 2^ logx = 1/4?

Nov 27, 2015

$x = \frac{1}{100}$ or $x = \frac{1}{e} ^ 2$ depending on whether $\log$ refers to ${\log}_{10}$ or ${\log}_{e}$.

#### Explanation:

We will use the following:

• ${\log}_{a} \left(a\right) = 1$

• $\log \left({a}^{x}\right) = x \log \left(a\right)$

• ${a}^{{\log}_{a} \left(x\right)} = x$

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

$\implies {\log}_{2} \left({2}^{\log} \left(x\right)\right) = {\log}_{2} \left({2}^{- 2}\right)$

$\implies \log \left(x\right) {\log}_{2} \left(2\right) = - 2 {\log}_{2} \left(2\right)$

$\implies \log \left(x\right) = - 2$

(Depending on the context, $\log$ may refer to ${\log}_{10}$ or ${\log}_{e}$. We will show both cases.)

$\implies {10}^{{\log}_{10} \left(x\right)} = {10}^{- 2}$ or ${e}^{{\log}_{e} \left(x\right)} = {e}^{- 2}$

$\implies x = {10}^{- 2} = \frac{1}{100}$ or $x = {e}^{-} 2 = \frac{1}{e} ^ 2$