Question

How do you solve `log(2x+1)+log(x-1) = log (4x-4)`?

Answer 1

`color(red)(x=3/2)`

Explanation:

`log(2x+1) + log(x-1) = log (4x-4)`

Recall that `log x + log y = log(xy)`, so

`log(2x+1)(x-1) = log(4x-4)`

Convert the logarithmic equation to an exponential equation.

`10^(log(2x+1)(x-1)) = 10^( log (4x-4))`

Remember that `10^logx =x`, so

`(2x+1)(x-1) = 4x-4`

Factor the right hand side.

`(2x+1)(x-1) = 4(x-1)`

Divide both sides by `x-1`, where `x≠1`.

`2x+1=4`

`2x=3`

`x=3/2`

Check:

`log(2x+1) + log(x-1) = log (4x-4)`

If `x=3/2`

`log(2(3/2)+1) + log(3/2-1) = log (4(3/2)-4)`

`log(3+1)+log(1/2) = log (6-4)`

`log4–log1-log2 = log 2`

`log(4/2) -0 = log 2`

`log2 = log2`

`x=3/2` is a solution.