Solve for x and check for any extraneous roots. log (x + 15) + log (x) = 2

We have the following logarithm rules:

1) log_c(a) + log_c(b) = log_c(ab)

2) log_c(x) = y => x = c^y

Given equation is:

log(x+15) + log(x) = 2

Apply the logarithm rule 1)

log[(x+15)x] = 2

=>log(x²+15x) = 2

We have log(x²+15x) = 2

Apply the logarithm rule 2), here c = 10 as nothing is mentioned

=>x²+15x = 10²

=>x²+15x-100 = 0

=>x²+20x-5x-100 = 0

=>x(x+20)-5(x+20) = 0

=>(x-5)(x+20) = 0

=>x-5 = 0; x+20 = 0

=>x = 5; x = -20

Check for extraneous roots by plugging these values in original equation

Check x = 5

log(5+15) + log(5) = 2

=>log(20) + log(5) = 2

=>log(20*5) = 2

=>log(100) = 2

=>100 = 10² , True

Check x = -20

log(-20+15) + log(-20) = 2

Here log (-x) is not defined and so log(-20) is not defined 

=> x= -20 is extraneous solution