#log_2(x+1)=log_4(x^2-x+4)# What is the value of x?

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Answer 1 · 2017-08-03T17:26:15

#x=1#

Equivalently we have

#(log_2(x+1))/(log_4 (x^2-x+4)) =( (log_e(x+1)/log_e 2))/( (log_e(x^2-x+4)/log_e4)) = 2(log_e(x+1)/ (log_e(x^2-x+4)))=1#

or

#2log_e(x+1)=log_e(x^2-x+4)# or

#(x+1)^2=(x^2-x+4)# or

#2x+1=-x+4# or

#3x=3# or #x = 1#