1. ## Log problem

Solve for x

loge(x)+loge(2x-1)=0

2. Dear smmmc,

I assume that you have written, $log_ex+log_e(2x-1)=0$

In that case,

$log_{e}x+log_{e}(2x-1)=0$

$\ln{x}=-\ln{(2x-1)}$

$\ln{x}=\ln(2x-1)^{-1}$

$x=\frac{1}{2x-1}$

$2x^2-x-1=0$

$x^2-\frac{1}{2}x-\frac{1}{2}=0$

$(x-\frac{1}{4})^2-\frac{1}{16}-\frac{1}{2}=0$

$(x-\frac{1}{4})^2=\frac{9}{16}$

$x=\pm{\frac{3}{4}}+\frac{1}{4}$

$x=1\mbox{ or }x=-\frac{1}{2}$

Hope this helps.

3. Originally Posted by smmmc
Solve for x

loge(x)+loge(2x-1)=0
$\ln{x} + \ln{(2x - 1)} = 0$

$\ln{[x(2x - 1)]} = 0$

$x(2x - 1) = e^0$

$x(2x - 1) = 1$

$2x^2 - x = 1$

$2x^2 - x - 1 = 0$

$2x^2 - 2x + x - 1 = 0$

$2x(x - 1) + 1(x - 1) = 0$

$(x - 1)(2x + 1) = 0$

$x - 1 = 0$ or $2x + 1 = 0$

$x = 1$ or $x = -\frac{1}{2}$.

4. wow thanks! could you clarify what |n is?

Thanks

5. Originally Posted by smmmc
wow thanks! could you clarify what |n is?

Thanks
$\ln$ stands for natural logarithm, and is alternative notation for $\log_e$.

6. Thanks!