How do you solve #log_2(x+2) + log_2(x+6)=5#?

1 Answer
A. S. Adikesavan · Shwetank Mauria
Apr 17, 2016

#x = 2#

Explanation:

Use #log a + log b = log ab# and, if #a^x=b, log_a(a^x)=x#,

#log_2 (x+2)+log_2(x+6)=log_2((x+2)(x+6))=5#, with #x+2 >0 and x+5 >0#. It is sufficient that #x> -2#.

So, #(x+2)(x+6)=2^5=32#
#x^2+8x-20=0#.
The roots are #x = 2# and #x = -10#.
As #x> -2#, the root #-10# is inadmissible. So, #x = 2#.

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