# How do you solve #log_2 x + log_2 (x + 2) = 3#?

##### 1 Answer

Use properties of logarithms and exponentiation to derive a quadratic equation, one of whose roots is a solution to the original equation, namely

#### Explanation:

If

Hence if

So we find:

#8 = 2^3 = 2^(log_2(x)+ log_2(x+2)) = 2^(log_2(x(x+2))) = x(x+2)#

Rearranging slightly, this becomes:

#0 = x^2+2x-8 = (x+4)(x-2)#

So

#log_2 2 + log_2(2+2) = 1 + 2 = 3#

#log_2(-4) + log_2(-4+2)#

#= (2+pi/ln(2) i)+ (1+pi/ln(2) i) = 3+(2 pi)/ln(2) i != 3#

So the only solution of the original equation is