1. Exponents as logs?

x^log x = 100x

How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks

2. Originally Posted by satx
x^log x = 100x

How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks
1. $\displaystyle x = 10^{\log(x)}$

$\displaystyle x^{\log( x)} = 100 \cdot x$

$\displaystyle \left(10^{\log(x)}\right)^{\log(x)} = 10^2 \cdot 10^{\log(x)}$

$\displaystyle \left(\log(x)\right)^2-\log(x)-2=0$

3. This is a quadratic equation in $\displaystyle \log(x)$ . Solve for $\displaystyle \log(x)$. You should come out with $\displaystyle \log(x) = 2 \vee \log(x)=-1$.

4. Calculate x.

3. Hello, satx!

Solve for $\displaystyle x\!:\;\;x^{\log x} \:=\: 100x$
Note that $\displaystyle x \neq 0.$

Divide by $\displaystyle x\!:\;\;x^{\log x - 1} \:=\:100$

$\displaystyle \text{Take logs, base }x\!:\;\;\log_x\left(x^{\log x - 1}\right) \:=\:\log_x(100) \quad\Rightarrow\quad (\log x - 1)\cdot\underbrace{\log_x(x)}_{\text{This is 1}} \:=\:\log_x(100)$

We have: . $\displaystyle \log x - 1 \:=\:\log_x100$

Change-of-base Formula: .$\displaystyle \log x - 1 \:=\:\frac{\log 100}{\log x} \quad\Rightarrow\quad \log x - 1\:=\:\frac{2}{\log x}$

Multiply by $\displaystyle \log x\!:\;\;(\log x)^2 - \log x \:=\:2 \quad\Rightarrow\quad (\log x)^2 - \log x - 2 \:=\:0$

Factor: .$\displaystyle (\log x + 1)(\log x - 2) \:=\:0$

And we have: .$\displaystyle \begin{array}{cccccccccccc} \log x + 1 \;=\:0 & \Rightarrow & \log x \:=\:\text{-}1 & \Rightarrow & x \:=\:10^{-1} \:=\:\boxed{\frac{1}{10}}\\ \\[-3mm] \log x - 2 \:=\:0 & \Rightarrow & \log x \:=\:2 & \Rightarrow & x \:=\:10^2 \:=\:\boxed{100}\end{array}$