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I needed help with the equation in the title, the base of $\log$ is $10$

I have tried the substitution of $x^2$ as $100$ and gotten $x= 10$ but I do not know how to solve it algebraically.

Robert Lee
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Lemun Juice
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  • Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Dec 13 '22 at 17:32
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    This equation does not have an algebraic solution in the ordinary sense of the word. You made a lucky guess. – Ethan Bolker Dec 13 '22 at 17:34
  • I got the lucky guess method , are you 100% sure about the no algebraic solution part? – Lemun Juice Dec 13 '22 at 18:04

1 Answers1

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Using logarithm properties you get $$ x^2 \log_{10}^5(x) = 10^2 \color{darkblue}{\implies} x^\frac{2}{5} \log_{10}(x) = 10^\frac{2}{5} \color{darkblue}{\implies}\log_{10}\left(x^{ x^\frac{2}{5} }\right) = 10^\frac{2}{5} \color{darkblue}{\implies} \color{green}{x}^{\color{green}{x}^{\frac{2}{5}}} = \color{green}{10}^{\color{green}{10}^{\frac{2}{5}} } $$ where by inspection you get $\color{green}{x} = \color{green}{10}$ as a solution.

By repeating the previous procedure, the general problem would in turn become solving $$ x^{x^{a}} = b $$ for some $a,b$ and $b>0$. For this, we can employ the Lambert-W function, which is defined to be such that if $xe^x =z$ then $x = W(z)$. We get $$ x^{x^{a}} = b \color{darkblue}{\implies} \color{purple}{a}\ln(x) e^{a\ln(x)} = \color{purple}{a}\ln(b)\color{darkblue}{\implies}a\ln(x) = W\left(a\ln(b) \right) \color{darkblue}{\implies} x= e^{\frac{W\left(a\ln(b) \right)}{a}} $$ In your case we would have $a= \frac{2}{5}$ and $b = 10^{{10}^{\frac{2}{5}}}$, thus \begin{align*} x = \exp\left[{\frac{W\left(\frac{2}{5}\ln\left(10^{{10}^{\frac{2}{5}}}\right) \right)}{\frac{2}{5}}} \right]= \exp\left[\frac{5}{2}W\left({10}^{\frac{2}{5}}\ln\left(10^{\frac25}\right) \right) \right] \end{align*} but since $W(xe^{x}) = x$ by definition of the Lambert-W, substituting $x=\ln(y)$ gives $W(y \ln(y)) = \ln(y)$, which allows us to simplify $$ x = \exp\left[\frac{5}{2}\ln\left(10^{\frac{2}{5}}\right) \right] = \exp[\ln(10)] = 10 $$ as expected.

Robert Lee
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