How do you solve log_(2x+1) N - log_4x = log_2 3log2x+1Nlog4x=log23?

1 Answer
Aug 18, 2016

x=1/2, N = 3 sqrt(2)/2x=12,N=322

Explanation:

log_(2x+1) N - log_4x = log_2 3log2x+1Nlog4x=log23

We know that

log_4x =log_e x/(log_e 2^2)=1/2log_e x/(log_e 2) = 1/2 log_2 xlog4x=logexloge22=12logexloge2=12log2x

so

log_(2x+1) N= log_2 3+1/2log_2 x=log_2(3 sqrt(x))log2x+1N=log23+12log2x=log2(3x)

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.

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