Logarithm Question (Find x)

Question

How to solve x for $$x^{2\log_{10}x}=\frac{x^5}{100}$$?

Answer 1

I'll write $\log$ for $\log_{10}$. Take the $\log$ of both sides. You get

$$2\log(x)^2 = 5\log(x) - 2.$$

Now let $y = \log(x)$. You have a quadratic equation $$ 2y^2 - 5y + 2 = 0 $$ which you can factor as $$ (2y - 1)(y-2) = 0. $$ So $y = 2$ or $y = 1/2$, so $x = 100$ or $x = \sqrt{10}$.

Answer 2

Starting with $$x^{2\log_{10}x}=\frac{x^5}{100}$$

Take $log_x$ of both sides of the equation.

Notice that: $log_{x}10 = \frac{1}{log_{10}x}$; and now solve!

Answer 3

Another easy way to solve . we can assume that $x^{2\log_{10}x}=x$ Then we find that$(2\log_{10}x}=1)$ Finally $x=\sqrt{10}$

Answer 4

An alternative basically identical to @hunter's solution would be to suppose a solution of the form $x=10^k$.

Substituting, we get:

$10^{2k^2}=10^{5k-2}$;

The transformation $10^{\cdot}$ is 1-1, so the solution for $k$ solves the quadratic $2k^2-5k+2=0$ with solutions $k=\frac{1}{2},2$.