# Solving equations with logarithms

I'm having trouble with solving equations that has logarithms in them. For example: $$x^{\log(x)} = \frac{100}{x}$$ How can I solve this? I have reed about how to do it but when I try to do the same with this example I just got stuck. The only thing I can think of doing is to draw the lines and see where they cross but I only get one answer $x=10$ but there should be two and I will need to show how I came up with the answer by showing calculations. Please help.

• Take the log of both sides. Apply the rules of logarithms and you wind up with a quadratic equation in the variable $\log x$. – David Mitra Feb 23 '14 at 15:05
• @DavidMitra but there is no log on the left side? – user3243600 Feb 23 '14 at 15:07
• $\log( x^{\log x})=\log(100/x)\Rightarrow \log x\cdot\log x=\log 100-\log x$. (Solving gives $x=10$ and $x= 1/100$.) – David Mitra Feb 23 '14 at 15:09

So you have $x^{\lg x}=\dfrac{100}{x}$. First of all, you have logarithms AND you have an exponential function. This power functions are ugly and you may remember about the special property of logarithm: $\log_a b^c=c\log_a b$ (there should be modulo, btw).
So you can apply a logarithm to both sides. What base — well, you already have a logarithm to base $10$ AND you have $100$ in the right side. So, $\lg$ is a good idea, indeed.
Now you have $\lg x\cdot\lg x=\lg 100 - \lg x$
That means $\lg^2x +\lg x-2=0$ — and you have quadratic equation.
The key is to take the log of both sides and then solve your equation: \begin{align*} \log x^{\log x}&=\log \frac{100}{x}\\ (\log x)(\log x)&=\log 100-\log x\\ (\log x)^2+\log x-\log 100&=0\\ \log x&=\frac{-1\pm\sqrt{1+8\log 10}}{2}\\ x&=e^\frac{-1\pm\sqrt{1+8\log 10}}{2} \end{align*}