How do you solve #log_7(2x^2 - 1) = log_7 4x#?

1 Answer
mason m
Nov 24, 2015

#2+sqrt6#

Explanation:

Note that if #log_ab=log_ac,# then #b=c#.

Because of this, we can say that #2x^2-1=4x#.

#2x^2-4x-1=0#

#x=(4+-sqrt(16+8))/2#

#x=2+-sqrt6#

However, the only answer is the positive one, #2+sqrt6#, sincein #log_ab#, #b>0#.

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