SOLUTION: Can anyone help with this? [In times the cuberoot of e^2] [log 125 to the base of 5] I did: [In(e^2)^1/3] [log 5^3 to the base of 5] [In e^2/3] [3log 5 to the base of 5] [1^

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Can anyone help with this? [In times the cuberoot of e^2] [log 125 to the base of 5] I did: [In(e^2)^1/3] [log 5^3 to the base of 5] [In e^2/3] [3log 5 to the base of 5] [1^      Log On


   

Question 22453: Can anyone help with this?
[In times the cuberoot of e^2] [log 125 to the base of 5]
I did:
[In(e^2)^1/3] [log 5^3 to the base of 5]
[In e^2/3] [3log 5 to the base of 5]
[1^2/3] [3(1)]
the cube root of 1^2(3)
the answer is 2, what did I do wrong and/or how do I finish?
thanks,
Sandy
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
You did fine.
You have [In e^2/3] [3log 5 to the base of 5]
ln e^(2/3) is (2/3)
3log 5 (base 5) is 3(1)=3
So you have (2/3)(3)= 2
Cheers,
Stan H.