Solve the equation and express the answer in terms of $\displaystyle e$.
$\displaystyle 7+9lnx=9$
This is what I am guessing:
$\displaystyle 9lnx=2$
$\displaystyle lnx={2/-9}$
$\displaystyle log_e(x)={2/-9}$
$\displaystyle log^{2/-9}=x$
Solve the equation and express the answer in terms of $\displaystyle e$.
$\displaystyle 7+9lnx=9$
This is what I am guessing:
$\displaystyle 9lnx=2$
$\displaystyle lnx={2/-9}$
$\displaystyle log_e(x)={2/-9}$
$\displaystyle log^{2/-9}=x$
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?