# Solving Logarithm Equation

Solve for x:$$x^2(\log_{10} x)^5 = 100$$ Here's what I've tried: $(\log_{10} x)^5 = A \\ \log_{ \log x} A= 5 = \frac{\log_{10} A}{\log_{10} \log x}$

Not sure how to continue

• The way you have written it's either transcendental (which it is either way) or there is some way to to randomly make sense of this problem. – Jared Oct 26 '16 at 4:47
• Let $\log_{10}x=y\implies x=10^y$ $$(10^y)^2y^5=100\iff y^5=10^{2-2y}$$ Clearly $yx=1$ is a solution and $$f(y)=y^5-10^{2-2y}$$ is increasing. – lab bhattacharjee Oct 26 '16 at 4:57

$$x^2(\log_{10}x)^{5}=10^2$$ $$x^{\frac{2}{5}}\log_{10}x=10^{\frac{2}{5}}$$ $$\log_{10}x^{x^\frac{2}{5}}=\log_{10}10^{10^\frac{2}{5}}$$ $$x^{x^\frac{2}{5}}=10^{10^\frac{2}{5}}$$ so the $$x=10$$
$$\log x = \left( \frac{10}{x} \right)^{2/5}$$
Now, $f(x) = \left( \frac{10}{x} \right)^{2/5}$ is a hyperbola with two branches. We are only concerned about values $x>0$ thus we only consider the right branch of this hyperbola. It is concave down and as $x \to \infty$ $f(x) \to 0$, thus it intesersect the log at one point only. By inspection, $x=10$ works.
Equations like this either have simple answers or ones that can only be found by numerical computation, so try the equivalent of the rational root theorem. If $x$ is rational, we need $\log_{10}x$ to be rational, which doesn't leave too many choices. You expect $\log_{10}x$ to be small, so ignore it (assume it is $1$). That gives $x=10$, which makes $\log{10}x=1$. Success.