Question 826046
The log2 bit means that the logs are taken to the base 2!
I normally write this down as log_2

In other words, if we have log_2(x) = p, which means that the log of x, taken to the base 2, is equal to p, then that is just another way of saying,
x = 2^p.
So now we have log_2(x-2) + log_2(x+5) = 1 
The thing about logs is that you can add them together, in fact they were, more less, designed for that to happen!
Adding together the logs, we get
log_2{(x-2)(x+5)} = 1
And the law of logs (one of them) tells us that
(x-2)(x+5) = 2^1 = 2
expanding the brackets,
x^2 + 3x - 10 = 2
x^2 + 3x - 12 = 0
Using the quadratic formula, {{{x = (-3 +- sqrt( 3^2+4*1*12 ))/(2*1) = (-3 +- sqrt( 9+48 ))/(2*1)}}}
So, the two solutions are: {{{x = (-3/2 + (1/2) sqrt( 57 ))}}}, {{{x = (-3/2 - (1/2) sqrt( 57 ))}}},