How do you solve #log_2(9x+5)=2+log_2(x^2-1)#?

1 Answer
Oct 16, 2016

Answer:

#x=-3/4 or3#

Explanation:

We need to write 2 in the form of #log_2?#
#2^2=4# so # log_2 4=2#

Rewrite the equation as
#log_2(9x+5)=log_2 4+log_2(x^2-1)#
Or using the addition rule of logs
#log_2(9x+5)=log_2 4(x^2-1)#
Therefore
#9x+5=4(x^2-1)#
Rearrange
#9x+5=4x^2-4#
#4x^2-9x-9=0#
Factorise
#(4x+3)(x-3)=0#
Thus #x=-3/4 or3#