Help Me, please

log (2x)  + log (x -5)    = 2

We can use this rule to simplify things, Manuel

log a  + log b  =  log (a * b)....so ....we can write the left sides as

log[ (2x * (x -5) ]  = 2

log [  2x^2 - 10x]  =  2      in exponential form we have

10^2  = 2x^2 - 10x

100  =  2x^2  - 10x        rearrange as

2x^2 - 10x - 100  = 0       divide through by 2

x^2 - 5x -50  =  0        factor as

(x - 10)(x + 5)  = 0

Setting each factor to 0 and solve for x  and we have that

x  = 10     or   x  = -5

We must reject  x  = -5   because it would mean that we would be taking logs of negative numbers in the original equation

So.....x  = 10