If #log_2(x−2)+log_2(x+4)=4#
then #x = 4#
as can be seen by replacing #x# with #4# in the sum from the original equation.
#log_2(4-2) + log_2(4+4)#
= # log_2(2) + log_2(8)#
Since #2^1 = 2# and #2^3 = 8#
(and therefore #log_2(2) = 1# and #log_2(8) = 3#
when x = 4,
#log_2(x−2)+log_2(x+4)= 1+3 = 4#
How to get there :
Let #2^j = x-2# (i.e. #j = log_2(x-2)#)
and
let #x^k = x+4# (i.e. #k = log_2(x+4)#)
From the given equation
#j + k = 4#
#(2^j) * (2^k) = (x-2)(x+4)#
combining, we get:
#2^(j+k) = x^2 + 2x - 8#
but, since #j+k = 4#
#2^4 = 16 = x^2 + 2x - 8#
therefore
#x^2 + 2x -24 = 0#
Simple factoring then gives
#(x - 4) (x + 6) = 0#
so
#x = 4# or #x = -6#
If #x = -6#
then
#log_2(x−2)# and #log_2(x+4)=4#
would be #log_2#'s of negative values (impossible).
Therefore
#x = 4#
