How to solve the equation $x^2(\log_{10} x )^5 = 100$?

Question

Find the value of $x$ (without hit and trial) from the equation $x^2(\log_{10} x )^5 = 100$.

Solving few steps I got $x= 10^{({\frac{x}{10}})^5} $.

Answer 1

$log_{10}x=y\\ \implies 10^y=x\\ \implies x^2=10^{2y}$

The equation will then become

$10^{2y}y^5=100$

One solution that I see is $y=1$, which will give $x=10$. Why do you wish to solve this equation without hit-and-trial method?

Answer 2

Substituting $x=10^y$ yields $(10^y)^2 y^5 = 100$. Obvious solution: $y=1$ or $x=10^1=10$.

Let $f(x)=x^2 (\log_{10} x)^5$. Then $f$ is strongly increasing function in $[0,+\infty)$ (as product of increasing functions). Therefore there exists only one solution and it is $x=10$.

Answer 3

You can solve it by looking at cases. Since $x$ appears in $log_{10}x$, we know $x \gt 0$. So we can divide by $x^2$. This gives $$(log_{10}x)^5 = \frac{100}{x^2}.$$

If $x \lt 10$, $LHS \lt 1$ and $RHS \gt 1$.

If $x = 10$, $LHS = RHS$.

If $x \gt 10$, $LHS \gt 1$ and $RHS \lt 1$.

Thus, the only solution is $x = 10$.

Answer 4

Just for fun. The solution of $$ x^2\big(\log_{10}(x)\big)^5 = M $$ in terms of the Lambert W function: $$ x = \frac{4\sqrt{10 M\;}(\ln 10)^{5/2}}{125\;W\left(\frac{2}{5}(\ln 10)\,M^{1/5}\right)^{5/2}} $$ to get all solutions, use all branches of W. Numerically, with $M=10$: $$ \vdots\\ -0.0031090972245807637012 - 0.0057756764312723178989 i\\ -0.0085157013468260475802 - 0.018595683863231787939 i\\ -0.070950674893432382960 - 0.16326665000276959405 i\\ 10.00000000000000000\\ -0.070950674893432382960 + 0.16326665000276959405 i\\ -0.0085157013468260475802 + 0.018595683863231787939 i\\ -0.0031090972245807637012 + 0.0057756764312723178989 i\\ \vdots $$