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

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?

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

Hint: Apply log on both side and try to solve

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.

• 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 \$. Commented May 17, 2013 at 17:32
• @RecklessReckoner thx, I had always thought that $\log x$ = $\ln x$ only happens in Mathematica or C++ xD
– arax
Commented May 17, 2013 at 17:42
• 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). Commented May 17, 2013 at 18:07

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

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 $logx = 1 \Rightarrow x = 10$

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 $x=10$

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.

• why prince Charles? Commented May 17, 2013 at 17:32
• @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. Commented May 17, 2013 at 17:35
• Yes, your original answer is correct if you use natural logarithms. (I did the same thing initially...) Commented May 17, 2013 at 18:24
• @RecklessReckoner: Thanks for the double-check. (+1 for your answer, by the way -- it's really the only one explaining the matter completely.) Commented May 17, 2013 at 18:45
• The issue was on my mind, as I work with students who are increasingly bumping into "log" not meaning the common logarithm. Commented May 17, 2013 at 19:04

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.

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.

• 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...) Commented May 17, 2013 at 19:15