If log_2(x−2)+log_2(x+4)=4log2(x−2)+log2(x+4)=4
then x = 4x=4
as can be seen by replacing xx with 44 in the sum from the original equation.
log_2(4-2) + log_2(4+4)log2(4−2)+log2(4+4)
= log_2(2) + log_2(8)log2(2)+log2(8)
Since 2^1 = 221=2 and 2^3 = 823=8
(and therefore log_2(2) = 1log2(2)=1 and log_2(8) = 3log2(8)=3
when x = 4,
log_2(x−2)+log_2(x+4)= 1+3 = 4log2(x−2)+log2(x+4)=1+3=4
How to get there :
Let 2^j = x-22j=x−2 (i.e. j = log_2(x-2)j=log2(x−2))
and
let x^k = x+4xk=x+4 (i.e. k = log_2(x+4)k=log2(x+4))
From the given equation
j + k = 4j+k=4
(2^j) * (2^k) = (x-2)(x+4)(2j)⋅(2k)=(x−2)(x+4)
combining, we get:
2^(j+k) = x^2 + 2x - 82j+k=x2+2x−8
but, since j+k = 4j+k=4
2^4 = 16 = x^2 + 2x - 824=16=x2+2x−8
therefore
x^2 + 2x -24 = 0x2+2x−24=0
Simple factoring then gives
(x - 4) (x + 6) = 0(x−4)(x+6)=0
so
x = 4x=4 or x = -6x=−6
If x = -6x=−6
then
log_2(x−2)log2(x−2) and log_2(x+4)=4log2(x+4)=4
would be log_2log2's of negative values (impossible).
Therefore
x = 4x=4
