SOLUTION: I need helping solving this for x: (ln x)^3=ln x^4 Can I rewrite it as (ln x)^3=4lnx ? Then can I divide both sides by lnx leaving (ln x)^2=4 ? Can I now square both sides leav

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I need helping solving this for x: (ln x)^3=ln x^4 Can I rewrite it as (ln x)^3=4lnx ? Then can I divide both sides by lnx leaving (ln x)^2=4 ? Can I now square both sides leav      Log On


   

Question 61154: I need helping solving this for x:
(ln x)^3=ln x^4
Can I rewrite it as (ln x)^3=4lnx ?
Then can I divide both sides by lnx leaving (ln x)^2=4 ?
Can I now square both sides leaving me with ln x=2 ?
Now I'm not sure what to do next.
Answer by mathick(4) About Me  (Show Source):
You can put this solution on YOUR website!
I need helping solving this for x:
(ln x)^3=ln x^4
Can I rewrite it as (ln x)^3=4lnx ?

Yes.

Then can I divide both sides by lnx leaving (ln x)^2=4 ?

Yes, but this assumes that you're not dividing both sides by 0, i.e. that ln(x) is not 0. This step wouldn't be valid in the case that ln(x) = 0, so this case (ln(x) = 0) needs to be treated separately.

Can I now square (root) both sides leaving me with ln x=2 ?

Right, ln (x) = 2, and also ln(x) = -2. (Taking the square root of both sides gives ln x = +2 and ln x = -2.)

Now I'm not sure what to do next

To solve ln(x) = 2 for x, exponentiate both sides:
e%5E%28ln%28x%29%29+=+e%5E2.

The left side simplifies, giving one of the final answers:

x+=+e%5E2.

The equation ln (x) = -2 can be solved similarly.

Finally, there is the case when ln(x) = 0. This happens when x = 1. To verify that this is a solution, you can plug it into the original equation and see if it checks out (gives a true equation).