# How to solve this logarithmic system of equations

I was trying to solve this system and I tried to express $$y = \frac{100}{x}$$ from the first equation and change into the second one and I got $$\frac{100}{x}\log_{10}{x} = 10$$ After some work I got to $$x = 10\log_{10}{x}$$ And I cannot solve this one.

$$\begin{cases}xy = 100 \\ y\log_{10}{x} = 10\end{cases}$$

Recall that there are log rules that may help. For instance, $$y\log_{10}x=\log_{10}x^y$$.

I think after employing this rule, you may be able to see the answer. What is $$\log_{10}x^y=10$$ actually saying, for instance?

Hint: From the second equation we get $$y=\frac{10}{\log_{10}{x}}$$ plugging this in the first one $$\frac{x}{\log_{10{x}}}=10$$ so $$x=10\log_{10}{x}$$ Here you will need a numerical method. From here we get the simpler equation $$10^{x/10}=x$$ With $$y=\frac{10}{\log{10}{x}}$$ we get

$$\frac{10x}{\log_{10}{x}}=100$$

• So it is basically $x^2=x^{10}$? – Mohammad Zuhair Khan Oct 28 '18 at 8:31
• @Raptor No. It's $x^2=10^x$ – yathish Oct 28 '18 at 8:37
• Oh, my bad! Transposition error. That means OP's $x=10\log_{10}x$ is wrong? – Mohammad Zuhair Khan Oct 28 '18 at 8:39
• Wouldn't you get $y=\frac{10}{log_{10}x}$ from the second equation? – Tartaglia's Stutter Oct 28 '18 at 8:43
• Yes this is true, just a typo! Thank you,corrected! – Dr. Sonnhard Graubner Oct 28 '18 at 8:47