SOLUTION: solve the equation log(x-6)+log(x-3)=1. List any extraneous roots.

Algebra.Com
Question 829074: solve the equation log(x-6)+log(x-3)=1. List any extraneous roots.
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
solve the equation log(x-6)+log(x-3)=1. List any extraneous roots.


10^1=(x-6)(x-3)
x^2-9x+18=10
x^2-9x+8=0
(x-1)(x-8)=0
x=1(reject,(x-3)>0))
or
x=8

RELATED QUESTIONS

Solve the equation and check your solutions. List any extraneous answer.... (answered by robertb)
solve log(x+2)+log x= 3 log 2 as algebraically and solve for any extraneous... (answered by nerdybill)
Solve the logarithmic equation log x + log (x-3)=... (answered by Edwin McCravy)
Solve the following. eliminate extraneous solutions. a 7^2x+1=4^x-2 b. 4^x= 3+18(4^-x) (answered by Alan3354)
Find the solution(s) of the equation?... (answered by user_dude2008)
find the solution(s) of the equation?... (answered by Earlsdon)
Solve the equation {{{log(16,(x))+log(8,(x))+log(4,(x))+log(2,(x))=6}}} (answered by jim_thompson5910)
Solve: Log x + log (x+6) =... (answered by jim_thompson5910)
solve the equation. log(x+2)= -log(x-1)... (answered by Alan3354)