# How do you solve logx^2 + logx^3 / log(100x) = 3?

Nov 3, 2015

The solutions are $x = {10}^{- 3} = \frac{1}{1000} = 0.001$ and $x = 10$.

#### Explanation:

First, use properties of logarithms to rewrite the equation $\log \left({x}^{2}\right) + \frac{\log \left({x}^{3}\right)}{\log \left(100 x\right)} = 3$ as $2 \log \left(x\right) + \frac{3 \log \left(x\right)}{\log \left(100\right) + \log \left(x\right)} = 3$, or

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

Next, you can multiply everything by $2 + \log \left(x\right)$ to get $4 \log \left(x\right) + 2 {\left(\log \left(x\right)\right)}^{2} + 3 \log \left(x\right) = 6 + 3 \log \left(x\right)$

Eliminate $3 \log \left(x\right)$ since it is found on both sides.

$4 \log \left(x\right) + 2 {\left(\log \left(x\right)\right)}^{2} = 6$

Divide all terms by $2$.

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

Subtract $3$ from both sides.

$2 \log \left(x\right) + {\left(\log \left(x\right)\right)}^{2} - 3 = 0$

Rewrite.

${\left(\log \left(x\right)\right)}^{2} + 2 \log \left(x\right) - 3 = 0$

Factor.

$\left(\log \left(x\right) + 3\right) \left(\log \left(x\right) - 1\right) = 0$.

Therefore, we seek values of $x$ so that $\log \left(x\right) = - 3$ and $\log \left(x\right) = 1$, giving $x = {10}^{- 3} = \frac{1}{1000} = 0.001$ and $x = 10$. You can check that these both work in the original equation.