solving logarithmic expression:$x= \frac{(\log_2(1+2x)) \cdot (1+2x)}{4}-5 $

Question

I need to solve logarithmic equation that contains a variable inside and outside the logarithm. I want to solve the following equation for x:

$$x= \frac{(\log_2(1+2x)) \cdot (1+2x)}{4}-5 $$

Note1: The base of logarithm is $2$.

Note2: The equation is supposed to be closed form.

Note3: The constant values are not important and can be changed.

Answer 1

$$ \underbrace{\log(2x+1)}_{g(x)} = \underbrace{4x+20\over 2x+1}_{f(x)}$$

Since $f$ is decreasing for $x>-{1\over 2}$ and $g$ is increasing the equation has at most one solution.

Draw both graphs and you will find an aproximate solution.

Answer 2

With a straightforward application of computer algebra in Mathematica, we find:

$$x =\frac{1}{2} \left(4 e^{W\left(\frac{9 \log (2)}{2}\right)}-1\right) \approx 5.33048.$$

where $W$ is the Lambert W function or ProductLog.

Answer 3

The equation is equivalent to the equation $f(x)=0$ where $$ f(x)=\log_2(1+2x)-\frac{20+4x}{1+2x} $$ where $x>-\frac12$. Note that $$ f'(x)=\frac{2+4x+36\log2}{(1+2x)^2\log2}>0, x>-\frac12 $$ Namely $f(x)$ is increasing when $x>-\frac12$. Since $f(0)=-5$ and $f(6)=0.315824$, $f(x)=0$ has a unique solution by the IVT.