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How do I find the solution to:

$$x^{\log(x)}=\frac{x^3}{100}$$

So I multiplied 100 both sides getting:

$$100x^{\log(x)}=x^3$$

Now what should I do?

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7 Answers 7

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Hint: Take the log of both sides. You will get a quadratic equation in $\log x$. The equation is even "nice."

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Hint: Apply log on both side and try to solve

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I suppose $\log$ means $\log_{10}$? I'm not familiar with this sort of notation. Take logarithm on both sides, and you will get $2+\log^2x=3\log x$. Substitute $\log x$ with t. And you get $t^2-3t+2=0$, therefore $(t-1)(t-2)=0$. That should do it.

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    $\begingroup$ Among most mathematicians, $ \ \log x \ $ means $ \ \log_e x \ $ or $ \ \ln x \ $. (To some, there is no other base worth talking about... :) ). Here, common logarithms would be notated as $ \ \log_{10} x \ $. $\endgroup$ Commented May 17, 2013 at 17:32
  • $\begingroup$ @RecklessReckoner thx, I had always thought that $\log x$ = $\ln x$ only happens in Mathematica or C++ xD $\endgroup$
    – arax
    Commented May 17, 2013 at 17:42
  • $\begingroup$ It occurs in those systems for exactly the reason I mention (consider who writes programming languages). Even going back to (gasp!) FORTRAN, the natural log function is written as LOG(X), while the common logarithm function is LOG10(X). $\endgroup$ Commented May 17, 2013 at 18:07
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$x^{\log(x)}=\frac{x^3}{100}$

Taking log on both sides you get :

$log(x) log(x) = log(\frac{x^3}{100})$ = log(x) log(x) = 3logx - 2log10 = 3logx -2

$\Rightarrow (log(x))^2 = 3logx -2 $

Now putting log(x) = t

$\Rightarrow t^2=3t-2$ Now you can solve for t as this is a quadratic in t. you get (t-2)(t-1) $\Rightarrow t = 2 ; t = 1$

$\Rightarrow logx = 2 \Rightarrow x = 100 $ ; and $ logx = 1 \Rightarrow x = 10$

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Since $(\log(x))^2=\log (x^{\log x})=\log (x^3/100)=3\log(x)-2$, we have $(\log(x))^2-3\log(x)+2=0$. Hence, $\log(x)=2$ and $\log(x)=1$. Therefore, $x=100$ atau $x=10$

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There are no solutions in the real numbers.


Edit: If, as RecklessReckoner suggests, the question was misstated and the intent was to use $\log_{10},$ then the solution can be found easily by taking logarithms.

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  • $\begingroup$ why prince Charles? $\endgroup$
    – beginner
    Commented May 17, 2013 at 17:32
  • $\begingroup$ @beginner: It's clear that the answer cannot be greater than $e^3$ because then the left side is greater than the right. For smaller $x$ it is easy to verify. $\endgroup$
    – Charles
    Commented May 17, 2013 at 17:35
  • $\begingroup$ Yes, your original answer is correct if you use natural logarithms. (I did the same thing initially...) $\endgroup$ Commented May 17, 2013 at 18:24
  • $\begingroup$ @RecklessReckoner: Thanks for the double-check. (+1 for your answer, by the way -- it's really the only one explaining the matter completely.) $\endgroup$
    – Charles
    Commented May 17, 2013 at 18:45
  • $\begingroup$ The issue was on my mind, as I work with students who are increasingly bumping into "log" not meaning the common logarithm. $\endgroup$ Commented May 17, 2013 at 19:04
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I am entering an answer just to point out a peculiarity of this equation, and the importance of interpreting the exponent "correctly". (This should also clarify Charles' answer.) I have graphed the quadratic functions $ \ (\ln x)^2 - (3 \ln x) + (\ln 100) \ $ in blue and $ \ (\log x)^2 - (3 \log x) + 2 \ $ in red. Notice that the blue curve has no x-intercepts; the quadratic equation for natural logarithms yields a negative discriminant since $ \ (-3)^2 < 4 \ln 100 \ $. So common logarithms are intended in this problem.

enter image description here

I had to use two graphs because the graph of the common logarithmic function is very shallow, but it does cross the x-axis at 100.

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  • $\begingroup$ An amusing variant of this problem occurred to me just now: for what base $ \ a \ $ does the equation $ \ x^{\log_a x} = \frac{x^3}{100} \ $ have exactly one solution? (We know it's somewhere between $ \ e \ $ and 10...) $\endgroup$ Commented May 17, 2013 at 19:15

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