Question

How do you solve `log_6(x + 19) + log_6(x) = 3`?

Answer 1

We must have `x>0` hence

#log_6(x + 19) + log_6(x) = 3=>log_6 x(x+19)=3=> x(x+19)=6^3=>x^2+19x-216=0=>x^2+27x-8x-216=0=> (x-8)*(x+27)=0=>x=8 or x=-27#

But `x=-27` is rejected so `x=8` is the only solution