# log

• Jan 15th 2008, 12:54 PM
rlarach
log
18. Solve the equation e^(2x) = 3

19. Solve the equation ln (3x - 2) + ln (x - 1) = 2 ln x.
• Jan 15th 2008, 02:11 PM
Pinsky
$\displaystyle e^{2x}=3$
then you ln the whole equation. And get
$\displaystyle x=\frac{1}{2}\ln{3}$

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

you solve the quadratic equation and get results.
• Jan 15th 2008, 05:05 PM
topsquark
Quote:

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

you solve the quadratic equation and get results.

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

-Dan
• Jan 16th 2008, 05:04 AM
Pinsky
See it now, tnx.