# Solve the equation 25log_2x=log_x2 ?

## $25 {\log}_{2} x = {\log}_{x} 2$ I have gotten to $\log {x}^{25} = \frac{1}{\log x}$ but I am not sure how to proceed from here. I am supposed to solve this without the use of a calculator, and the answer is given as $x = {2}^{\left(\frac{+}{-}\right) \frac{1}{5}}$. How should I go about solving this?

Apr 22, 2018

I tried this:

#### Explanation:

I would try changing the base of the second log as:

${\log}_{x} 2 = {\log}_{2} \frac{2}{\log} _ 2 \left(x\right) = \frac{1}{\log} _ 2 \left(x\right)$

so we get:

$25 {\log}_{2} x = \frac{1}{\log} _ 2 \left(x\right)$

rearrange:
$25 {\log}_{2} x \cdot {\log}_{2} x = 1$
$25 {\left[{\log}_{2} x\right]}^{2} = 1$
${\left[{\log}_{2} x\right]}^{2} = \frac{1}{25}$
take the square root of both sides to get rid of the square:
${\log}_{2} x = \pm \frac{1}{5}$
apply the definition of log:
$x = {2}^{\pm \frac{1}{5}}$

Apr 22, 2018

$x = \frac{1}{2} ^ \left(\frac{1}{5}\right)$, $x = {2}^{\frac{1}{5}}$

#### Explanation:

$25 {\log}_{2} \left(x\right) = {\log}_{x} \left(2\right)$
${\log}_{2} \left({x}^{25}\right) = {\log}_{x} \left(2\right)$
$\log \frac{{x}^{25}}{\log} \left(2\right) = \log \frac{2}{\log} \left(x\right)$
$25 \log \left(x\right) \cdot \log \left(x\right) = \log \left(2\right) \cdot \log \left(2\right)$
$25 {\log}^{2} \left(x\right) = {\log}^{2} \left(2\right)$
$5 \log \left(x\right) = \pm \log \left(2\right)$
$\log \left({x}^{5}\right) = \pm \log \left(2\right)$

$\log \left({x}^{5}\right) = \log \left({2}^{- 1}\right) = \log \left(\frac{1}{2}\right)$
${x}^{5} = \frac{1}{2}$
$x = \frac{1}{2} ^ \left(\frac{1}{5}\right)$

$\log \left({x}^{5}\right) = \log \left(2\right)$
${x}^{5} = 2$
$x = {2}^{\frac{1}{5}}$