Solve $(\log_5 x)^2 – 3\log_5(x) + 2 = 0$.
I tried to solve this question but got stuck:
\begin{align}
(\log_5 x)^2 – 3\log_5(x) + 2 &= 0 \\
(\log_5 x)^2 &= \log_5(x^3) – 2 \\
\log_{(\log_5 x)}(\log_5 x)^2 &= \log_{( \log_5 x)}(\log_5(x^3) – \log_5 25) \\
2 &= \log_{(\log_5 x)}\left( \log_5\frac{x^3}{25} \right)
\end{align}
Was my approach wrong? Where do I go from here?
Best Answer
This is just a quadratic equation, but instead of variables you have logarithms. You can let $u=\log_5x$ and get
$$u^2-3u+2=0$$
$$(u-2)(u-1)=0$$
from which $\log_5x=1$ and $\log_5x=2$.