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

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Answer 1 · 2015-08-01T19:51:28

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

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

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.

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