SOLUTION: log(2x-3)+ log(x-2)=2logx Please solve for x and show the work thank you

Click here to see ALL problems on logarithm

Question 311717: log(2x-3)+ log(x-2)=2logx
Please solve for x and show the work thank you
Found 2 solutions by ankor@dixie-net.com, scutechandni10:
Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website!
log(2x-3)+ log(x-2) = 2log(x)
:
log((2x-3)*(x-2)) = log(x^2)
FOIL
log(2x^2-4x-3x+6) = log(x^2)
:
log(2x^2 - 7x + 6) = log(x^2)
therefore
2x^2 - 7x + 6 = x^2
:
2x^2 - x^2 - 7x + 6 = 0
:
x^2 - 7x + 6 = 0
Factor
(x - 6)(x - 1) = 0
Two solutions
x = 6
and
x = 1
;
:
Check both solutions in original problem
log(2x-3)+ log(x-2) = 2log(x)
x=6
log(2(6)-3)+ log(6-2) = 2log(6)
log(9) + log(4) = log(6^2)
log(36) = log(36)
x = 6; a good solution
:
x=1, not a solution, the log of a negative number not permitted

Answer by scutechandni10(7) (Show Source):
You can put this solution on YOUR website!
log(2x-3)+ log(x-2) = 2log(x)
log((2x-3)*(x-2)) = log(x^2)
log(2x^2-4x-3x+6) = log(x^2)
log(2x^2 - 7x + 6) = log(x^2)
Solve quadratic equation;
therefore
2x^2 - 7x + 6 = x^2
2x^2 - x^2 - 7x + 6 = 0
x^2 - 7x + 6 = 0
Factors:
(x - 6)(x - 1) = 0
Either x=6 or x=1
Check both solutions (x=1 or x=6) in
log(2x-3)+ log(x-2) = 2log(x) by substituting value of x
By using x=6
log(2(6)-3)+ log(6-2) = 2log(6)
log(9) + log(4) = log(6^2)
log(36) = log(36)
By using x=1
log(2(1)-3)+log(1-2)=2log(1)
log(-1)+log(-1)=2log1
Apply law of logarithm of a product
log(-1*-1)=log1^2
log(1)=log1^2
They are not equal
Therefore; x=1, not a solution
So the value of x in the question given will be 6 (x=6)
Good luck!