SOLUTION: logx+log3-(log3x-1)=log4

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.

RELATED QUESTIONS
Solve for X:... (answered by lwsshak3)

log3x+log3(x-2)=1 (answered by ewatrrr)

log3x=log4+log(x-1) solve this logarithmic... (answered by Susan-math)

log3-logx (answered by NAMath)

1/3 log3 x^6 +1/6 log3 x^6 - 1/9 log3x (answered by Theo)

How do I solve the following problem: logx-log3=log4-log(x+4). (answered by sdmmadam@yahoo.com)

Please help me. thanks 1.Solve for x: log4x+log4(x+4)=9 2. For what value of x (answered by robertb)

log3(3x)=log3x+log3(4-x) (answered by user_dude2008)

... (answered by lwsshak3)