# How do you solve log_(2x+1) N - log_4x = log_2 3?

Aug 18, 2016

$x = \frac{1}{2} , N = 3 \frac{\sqrt{2}}{2}$

#### Explanation:

${\log}_{2 x + 1} N - {\log}_{4} x = {\log}_{2} 3$

We know that

${\log}_{4} x = {\log}_{e} \frac{x}{{\log}_{e} {2}^{2}} = \frac{1}{2} {\log}_{e} \frac{x}{{\log}_{e} 2} = \frac{1}{2} {\log}_{2} x$

so

${\log}_{2 x + 1} N = {\log}_{2} 3 + \frac{1}{2} {\log}_{2} x = {\log}_{2} \left(3 \sqrt{x}\right)$

Now making equivalents of argument and basis

{(N = 3 sqrt(x)), (2x+1=2):}

Solving for $N , x$ we get at

$x = \frac{1}{2} , N = 3 \frac{\sqrt{2}}{2}$ which is a solution.

Of course there are infinite solutions according to the attached figure in which appear the solution points for $N$ vertical and $x$ horizontal.