# solve log equation

• May 6th 2008, 08:26 PM
theevilp0ptart
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\$
• May 6th 2008, 08:42 PM
blair_alane
Quote:

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!
• May 6th 2008, 08:48 PM
o_O
Quote:

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?
• May 7th 2008, 05:16 AM
theevilp0ptart
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 AM
Tangera