Question

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

Answer 1

Use properties of logarithms and exponentiation to derive a quadratic equation, one of whose roots is a solution to the original equation, namely `x=2`

Explanation:

If `a, b, c > 0` then `log_c(a) + log_c(b) = log_c(ab)`

Hence if `x > 0` then `log_2(x) + log_2(x+2) = log_2(x(x+2))`

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 `x=-4` or `x=2`

`x=2` is a solution of the original equation, as we can check if we wish:

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

`x=-4` does not satisfy `x > 0` so we are not guaranteed that it will work, even if we allow Complex logarithms. In fact we find:

`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 `x=2`