SOLUTION: The excersise goes like that: ''Solve this logarithmix equation: 1. 2*log*(2x-1) - log*(3x-2x^2) = log*(4x-3)-log*x Thank you for replying!

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: The excersise goes like that: ''Solve this logarithmix equation: 1. 2*log*(2x-1) - log*(3x-2x^2) = log*(4x-3)-log*x Thank you for replying!      Log On


   

Question 1028182: The excersise goes like that:
''Solve this logarithmix equation:
1. 2*log*(2x-1) - log*(3x-2x^2) = log*(4x-3)-log*x
Thank you for replying!
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
log%28%282x-1%29%5E2%29-log%283x-2x%5E2%29=log%28%284x-3%29%2F%28x%29%29

log%28%282x-1%29%5E2%2F%283x-2x%5E2%29%29=log%28%284x-3%29%2Fx%29

Temporary change in variables saying as log%28%28A%29%29=log%28%28B%29%29.


%282x-1%29%5E2%2F%283x-2x%5E2%29=%284x-3%29%2Fx

x%282x-1%29%5E2=%283x-2x%5E2%29%284x-3%29

x%284x%5E2-4x%2B1%29=12x%5E2-9x-8x%5E3%2B6x%5E2

4x%5E3-4x%5E2%2Bx=18x%5E2-9x-8x%5E3

12x%5E3-22x%5E2%2B10x=0

6x%5E3-11x%5E2%2B5x=0

x%286x%5E2-11x%2B5%29=0
You are not interested in x=0, because you cannot have log of negative values as would happen in the original equation. Put attention instead on the remaining roots of the quadratic factor.

Discrim, %28-11%29%5E2-4%2A6%2A5=121-6%2A20=1
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x=%2811%2B-+1%29%2F12
highlight%28system%28x=5%2F6%2Cor%2Cx=1%29%29