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.

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