# solve log equation

• May 6th 2008, 09:26 PM
theevilp0ptart
solve log equation
Solve the equation and express the answer in terms of $e$.

$7+9lnx=9$

This is what I am guessing:
$9lnx=2$
$lnx={2/-9}$
$log_e(x)={2/-9}$
$log^{2/-9}=x$
• May 6th 2008, 09:42 PM
blair_alane
Quote:

Originally Posted by theevilp0ptart
Solve the equation and express the answer in terms of $e$.

$7+9lnx=9$

This is what I am guessing:
$9lnx=2$
$lnx={2/-9}$
$log_e(x)={2/-9}$
$log^{2/-9}=x$

why would it be -9? it was positive just a step ago!
• May 6th 2008, 09:48 PM
o_O
Quote:

Originally Posted by theevilp0ptart
Solve the equation and express the answer in terms of $e$.

$7+9lnx=9$

This is what I am guessing:
$9lnx=2$
$lnx={2/{\color{red}-}9}$
$log_e(x)={2/{\color{red}-}9}$
${\color{blue}log^{2/{\color{red}-}9}=x}$

NO!

"log" is not a variable. Just like you don't say "sin^2". It has to be the sin OF something. Similarly, you have to take the LOGARITHM of something.

Recall the definition: $log_{a} b = c \: \: \iff \: \: a^{c} = b$

The log operator takes the form of $a^{c} = b$ and allows you to solve for the exponent c if you have a and b. Look at the definition and see how it works here.

And as blair_alane said, where did the negative come from?
• May 7th 2008, 06:16 AM
theevilp0ptart
So would it be this:

$9lnx=2$
$lnx=2/9$
$log_{e}x=2/9$
$e^{2/9}=x$
• May 7th 2008, 06:44 AM
Tangera