# Thread: 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 $\displaystyle \frac{e}{3}$. 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: $\displaystyle x=1,3$

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.

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

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

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

Central solution should be $\displaystyle x=1.$

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

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

5. Originally Posted by colby2152
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: $\displaystyle x=-3,0,3$
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

Sorry about that!

7. I was even thinking that when I did the problem, but I still kept x = 0 as a solution. To the OP, you cannot take the natural log of a number zero or less.