Question 1071648: logx+log3-(log3x-1)=log4
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! what i think you are asking is:
log(x) + log(3) - log(3x-1) = log(4)
if so, you would do the following:
subtract log(3) from both sides of the equation to get:
log(x) - log(3x-1) = log(4) - log(3)
since log(a) - log(b0 = log(a/b), this equation becomes:
log(x/(3x-1)) = log(4/3)
since log(a) = log(b) is true if and only if a = b, this equation becomes:
x/(3x-1) = 4/3
cross multiply to get:
3x = 4 * (3x-1)
sinmplify to get:
3x = 12x - 4
subtract 3x from both sides of the equation and add 4 to both sides of the equation to get:
4 = 9x
divide both sides of the equation by 9 and solve for x to get:
x = 4/9
that's your solution.
to confirm it's a good solution, replace x in the original equation with it and evaluate that equation to see if it's true.
it's true if the left side of the equation equals the right side of the equation.
start with:
log(x) + log(3) - log(3x-1) = log(4)
replace x with 4/9 to get:
log(4/9) + log(3) - log(3*4/9 - 1) = log(4)
evaluate to get:
.6020599913 = .6020599913
since this equation is true, the solution is good.
your solution is that x = 4/9.
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