Answer to Question #194513 in Algebra for Rebira

Question

X^{log x}=100x

Answer 1

As the usage of $\log x$ requires $x\neq 0$, we can divide both sides by $x$ knowing that it is non-zero.

$x^{\log x}/x = 100$

As $a^b/a^c = a^{b-c}$ and $a=a^1$, we have

$x^{\log x}/x = x^{\log(x)-1}$

$x^{\log (x)-1}=100$

As we can write $a^b=10^{b\log a}$, we have

$10^{(\log(x)-1)\cdot\log(x)}=100$

Finally, as $100=10^2$ we conclude

$(\log(x)-1)\cdot\log(x) = 2$

$\log^2(x)-\log(x)-2=0$

Solving this equation quadratic in $\log(x)$ gives us

$\begin{cases} \log(x_1)=2 \\ \log(x_2)=-1 \end{cases}$

And by removing the log we obtain two solutions :

$\begin{cases} x_1=100 \\ x_2=0.1 \end{cases}$