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

$\displaystyle 7+9lnx=9$

This is what I am guessing:

$\displaystyle 9lnx=2$

$\displaystyle lnx={2/-9}$

$\displaystyle log_e(x)={2/-9}$

$\displaystyle log^{2/-9}=x$

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- May 6th 2008, 08:26 PMtheevilp0ptartsolve log equation
Solve the equation and express the answer in terms of $\displaystyle e$.

$\displaystyle 7+9lnx=9$

This is what I am guessing:

$\displaystyle 9lnx=2$

$\displaystyle lnx={2/-9}$

$\displaystyle log_e(x)={2/-9}$

$\displaystyle log^{2/-9}=x$ - May 6th 2008, 08:42 PMblair_alane
- May 6th 2008, 08:48 PMo_O
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? - May 7th 2008, 05:16 AMtheevilp0ptart
So would it be this:

$\displaystyle 9lnx=2$

$\displaystyle lnx=2/9$

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

$\displaystyle e^{2/9}=x$ - May 7th 2008, 05:44 AMTangera
Hello! Yup, your answer is correct. :)