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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.

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

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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?

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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

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

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

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