SOLUTION: 2 loga X + loga 2 = loga (5x+3)
I thought:
X^2 + 2 = 5X+3
X^2 -5X - 1 = 0
The answer is suppose to be 3. I don't understand how to get there since it doesn't factor.
Question 24227: 2 loga X + loga 2 = loga (5x+3)
I thought:
X^2 + 2 = 5X+3
X^2 -5X - 1 = 0
The answer is suppose to be 3. I don't understand how to get there since it doesn't factor. Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! SEE MY COMMENTS BELOW
2 loga X + loga 2 = loga (5x+3)..ALL ARE TO SAME BASE A . SO WE CAN DROP THE BASE.
I thought:
X^2 + 2 = 5X+3..YOU DID 2*LOG X = LOG (X^2)..GOOD..BUT BEFORE TAKING ANTILOGS ,YOU SHOULD BRING L.H.S. AND R.H.S. TO SINGLE LOG TERM.R.H.S. HAS ONLY ONE LOG TERM LOG (5X+3)..IT IS OK .BUT L.H.S .HAS 2 LOG TERMS .............
LOG (X^2)+LOG 2..FIRST YOU SHOULD COMBINE THEM INTO ONE ..USING....
LOG X +LOG Y = LOG (X*Y).SO WE GET L.H.S. AS
LOG (X^2)+LOG 2 = LOG (2*X^2),SO THAT WE HAVE LOG (2*X^2)=LOG (5X+3)..NOW WE CAN TAKE ANTILOGS.TO GET 2*X^2=5X+3
2X^2-5X-3=0
2X^2-6X+X-3=0..(I THINK YOU KNOW THIS .OTHERWISE COME BACK & I SHALL EXPLAIN)
2X(X-3)+1(X-3)=0
(X-3)(2X+1)=0
HENCE X-3 = 0 ...OR...X=3
THE OTHER ALTERNATIVE 2X+1=0..GIVES...2X=-1...OR X=-1/2..WHICH LEADS TO LOG OF A NEGATIVE NUMBER AND HENCE IS NOT PERMISSIBLE.
HENCE X=3
answer is suppose to be 3. I don't understand how to get there since it doesn't factor.
