How do you solve log_2 (x + 3) - log_2 (x – 4) = log_2 8?

1 Answer
Mar 25, 2016

x=5

Explanation:

Since all the logs have the same base you can take advantage of the property: log_n(a/b)= log_n(a) - log_n(b)

Rewrite the equation:

log_2((x+3)/(x-4)) = log_2(8)

Rewrite according to: log_b(x)=y, b^y = x

2^(log_2(8)) = (x+3)/(x-4)

Note that a^(log_a(b)) = b, so you can rewrite as:

8 = (x+3)/(x-4)
Solve for x:
8(x-4) = (x+3)
8x-32 = x+3
7x-32=3
7x=35
x = 5