# Math Help - Solving Log Equations

1. ## Solving Log Equations

1.) xe^-2x + 2e^-2 = 0

2.) x^2 ln x - 9 ln x = 0

3.) 2 ln 3x - 1 = 1

Thanks in advance for any help!

2. The answer to the last one is $\frac{e}{3}$. Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that?

As far as the first two are concerned...

2) $(x^2-9)ln(x)=0$

Solve for x: $x=1,3$

3. No3.

$

\begin{gathered}
2\ln 3x - 1 = 1 \\
\ln (3x)^2 - \ln (e) = \ln (e) \\
\frac{{9x^2 }}
{e} = e \\
x = \frac{e}
{3} \\
\end{gathered}

$

No2.

$
\begin{gathered}
x^2 \ln x - 9\ln x = 0 \\
(x^2 - 9)\ln x = \ln 1 \\
\ln x^{(x^2 - 9)} = \ln 1 \\
x^{(x^2 - 9)} = 1 \\
x = 1,3\\
\end{gathered}

$

4. Originally Posted by colby2152
2) $(x^2-9)ln(x)=0$

Solve for x: $x=-3,0,3$
Little mistake here

Central solution should be $x=1.$

Originally Posted by toop
3.) 2 ln 3x - 1 = 1
This is $\ln3x=1.$ Can you take it from there?

Anyway use parenthesis please, it could be $2\ln(3x-1)=1.$ (See my signature.)

5. Originally Posted by colby2152
The answer to the last one is $\frac{e}{3}$. Add the one to the right side; divide by two, and then exponentiate each side. Do you follow that?

As far as the first two are concerned...

2) $(x^2-9)ln(x)=0$

Solve for x: $x=-3,0,3$
Hi,

only a short notice: The ln()-function is not definde for negative numbers. Therefore x = -3 is not a solution of this equation.

6. Oops! I posted the problem wrong for #3... It should have been

3.) 2 ln 3x = 1