2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jared Oct 26 '16 at 4:47
  • $\begingroup$ 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. $\endgroup$ – lab bhattacharjee Oct 26 '16 at 4:57
1
$\begingroup$

$$x^2(\log_{10}x)^{5}=10^2$$ $$x^{\frac{2}{5}}\log_{10}x=10^{\frac{2}{5}}$$ $$\log_{10}x^{x^\frac{2}{5}}=\log_{10}10^{10^\frac{2}{5}}$$ $$x^{x^\frac{2}{5}}=10^{10^\frac{2}{5}}$$ so the $$x=10$$

$\endgroup$
1
$\begingroup$

Rewrite your expression as follows

$$ \log x = \left( \frac{10}{x} \right)^{2/5} $$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.