1. solve log equation

Solve the equation and express the answer in terms of $\displaystyle e$.

$\displaystyle 7+9lnx=9$

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

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

$\displaystyle 7+9lnx=9$

This is what I am guessing:
$\displaystyle 9lnx=2$
$\displaystyle lnx={2/-9}$
$\displaystyle log_e(x)={2/-9}$
$\displaystyle log^{2/-9}=x$
why would it be -9? it was positive just a step ago!

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

$\displaystyle 7+9lnx=9$

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

The log operator takes the form of $\displaystyle a^{c} = b$ 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?

4. So would it be this:

$\displaystyle 9lnx=2$
$\displaystyle lnx=2/9$
$\displaystyle log_{e}x=2/9$
$\displaystyle e^{2/9}=x$