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

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Answer 1 · 2016-08-18T09:44:57

$x=1/2, N = 3 sqrt(2)/2$

$log_(2x+1) N - log_4x = log_2 3$

We know that

$log_4x =log_e x/(log_e 2^2)=1/2log_e x/(log_e 2) = 1/2 log_2 x$

so

$log_(2x+1) N= log_2 3+1/2log_2 x=log_2(3 sqrt(x))$

Now making equivalents of argument and basis

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

Solving for $N, x$ we get at

$x=1/2, N = 3 sqrt(2)/2$ 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.

enter image source here

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