Thread: 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!

The answer to the last one is . 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)

Solve for x:

No3.



No2.

Originally Posted by colby2152

2)

Solve for x:

Little mistake here

Central solution should be

Originally Posted by toop

3.) 2 ln 3x - 1 = 1

This is Can you take it from there?

Anyway use parenthesis please, it could be (See my signature.)

Originally Posted by colby2152

The answer to the last one is . 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)

Solve for x:

Hi,

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

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

3.) 2 ln 3x = 1

Sorry about that!

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.