Given

$\displaystyle \int \frac{\ln bx}{x^2} ~dx =-\frac{\ln bx}{x}-\frac{1}{x}+C $

Then find $\displaystyle b$ where

$\displaystyle \int_{\frac{1}{b}}^4 \frac{5\ln bx}{x^2} ~dx = 5$

I'm saying that

$\displaystyle 5\int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 5$

Dividing both sides by $\displaystyle 5$

$\displaystyle \int_{\frac{1}{b}}^4 \frac{\ln bx}{x^2} ~dx = 1$

$\displaystyle \left[ -\frac{\ln bx}{x}-\frac{1}{x}\right]_{\frac{1}{b}}^4 = 1$

$\displaystyle \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -\frac{\ln b\frac{1}{b}}{\frac{1}{b}}-\frac{1}{\frac{1}{b}}\right] = 1$

$\displaystyle \left[ -\frac{\ln 4b}{4}-\frac{1}{4}\right]-\left[ -0-b\right] = 1$

$\displaystyle -\frac{\ln 4b}{4}-\frac{1}{4}+b = 1$

Here is where I am stuck because I can't find an explicit way to solve for $\displaystyle b$ Am I missing a log law or something similar?

Thanks..