1. ## log

18. Solve the equation e^(2x) = 3

19. Solve the equation ln (3x - 2) + ln (x - 1) = 2 ln x.

2. $e^{2x}=3$
then you ln the whole equation. And get
$x=\frac{1}{2}\ln{3}$

$ln{(3x-2)}+ln{(x-1)}=2\ln{x}$
$ln{(3x-2)(x-1)}=\ln{x^2}$
$(3x-2)(x-1)=x^2$
$2x^2-5x+2=0$
$x_1=2$
$x_2=\frac{1}{2}$

you solve the quadratic equation and get results.

3. Originally Posted by Pinsky
$ln{(3x-2)}+ln{(x-1)}=2\ln{x}$
$ln{(3x-2)(x-1)}=\ln{x^2}$
$(3x-2)(x-1)=x^2$
$2x^2-5x+2=0$
$x_1=2$
$x_2=\frac{1}{2}$

you solve the quadratic equation and get results.
That's a nice derivation. But did you notice that $x_2 = \frac{1}{2}$ is not a solution? It doesn't lie in the domain of the original expression.

-Dan

4. See it now, tnx.