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!
The answer to the last one is $\displaystyle \frac{e}{3}$. 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) $\displaystyle (x^2-9)ln(x)=0$
Solve for x: $\displaystyle x=1,3$
No3.
$\displaystyle
\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.
$\displaystyle
\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}
$