How to solve this logarithmic system of equations

Question

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.

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

score 2 · Answer 1 · answered Oct 28 '18 at 08:35

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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?

answered Oct 28 '18 at 08:35

Tartaglia's Stutter

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Dr. Sonnhard Graubner · Answer 2 · 2018-10-28T08:55:40.147

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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

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

edited Oct 28 '18 at 08:55

answered Oct 28 '18 at 08:18

Dr. Sonnhard Graubner

95,283

So it is basically $x^2=x^{10}$? – For the love of maths Oct 28 '18 at 08:31
@Raptor No. It's $x^2=10^x$ – yathish Oct 28 '18 at 08:37
Oh, my bad! Transposition error. That means OP's $x=10\log_{10}x$ is wrong? – For the love of maths Oct 28 '18 at 08:39
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Wouldn't you get $y=\frac{10}{log_{10}x}$ from the second equation? – Tartaglia's Stutter Oct 28 '18 at 08:43
Yes this is true, just a typo! Thank you,corrected! – Dr. Sonnhard Graubner Oct 28 '18 at 08:47
Then, plugging that into the first equation, I think all of your $2$s should be $10$s. Ultimately you wind up at $x=10\log_{10}x$, where the OP had reached. – Tartaglia's Stutter Oct 28 '18 at 08:50
Yes $$100/10=10$$ oh man,sorry thousend times – Dr. Sonnhard Graubner Oct 28 '18 at 08:56

How to solve this logarithmic system of equations

2 Answers2