Solve the equation (logx)^2-4logx=0?

Jul 6, 2017

See below.

Explanation:

${\left(\log x\right)}^{2} - 4 \log x = 0$ or

$\left(\log x - 4\right) \log x = 0$ so we have the solutions

$\left\{\begin{matrix}\log x - 4 = 0 \to x = {e}^{4} \\ \log x = 0 \to x = 1\end{matrix}\right.$

Jul 6, 2017

$x = 1$ or $x = {e}^{4}$

Explanation:

We have:

${\left(\ln x\right)}^{2} = \ln \left({x}^{4}\right)$

Using the rules of logarithms, this can be written as:

${\left(\ln x\right)}^{2} = 4 \ln x$

$\therefore {\left(\ln x\right)}^{2} - 4 \ln x = 0$
$\therefore \ln x \left(\ln x - 4\right) = 0$

This would lead to two possible solutions:

Either:

$\ln x = 0 \implies x = {e}^{0}$
$\text{ } \implies x = 1$

Or:

$\ln x - 4 = 0 \implies \ln x = 4$
$\text{ } \implies x = {e}^{4}$