Please tell me if this is right...

Solve the logarithmic equation:

log_4(x-18) + log_4 (x-3) = 2
log_4((x-18)(x-3)=4^2
x^2-21x+54=16
x^2-21x+38=0
(x-2)(x-19)=0
x=2, x=19

Question

Please tell me if this is right...

Solve the logarithmic equation:

log_4(x-18) + log_4 (x-3) = 2
log_4((x-18)(x-3)=4^2
x^2-21x+54=16
x^2-21x+38=0
(x-2)(x-19)=0
x=2, x=19 <--- Answer

hartnn · Answer

your final answer is correct but the 2nd step will be 
log_4((x-18)(x-3)=2
then 3rd step x^2-21x+54=16 is correct

hartnn · Answer

or your 2nd step could be 
log_4((x-18)(x-3)=log_4 (4^2)

anonymous · Answer

yea I did that but I skipped the step on here. I wrote that on my paper.

hartnn · Answer

then you are correct

hartnn · Answer

oh, wait i just realized, x=2 cannot be the answer

hartnn · Answer

because that would make x-3 as negative
then log (x-3) will not be defined @Fakshon