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^

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) (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.
RELATED QUESTIONS
1.Write as a difference of logarithms a) log(base 4)11 b) log(base 3)12 c) log(base... (answered by MathLover1)

Could you please help me with this problem Use: log of 3 in base x = .275 (answered by venugopalramana)

This is the question: log(base 5) 125^1/2 + log(base 11) 121^1/3= 13/6 What I tried: (answered by nerdybill)

Please help me solve this equation: Solve the logarithmic equation logarithmic... (answered by jim_thompson5910)

Write in terms of the NATURAL LOG. (ln) 1. log (base 4)x 2. log (base 10) (x+1) 3. log (answered by solver91311,jsmallt9)

1. Solve the following equations a) log(base 5)x+2 + log(base 5) x-2 - log(base 5)4 =... (answered by mananth)

Solve the equation for x: 2log base 5 of x + log base 5 of 3 = log base 5(1/125) (answered by lwsshak3)

How do i find the nth term for these sequences? a) log [base 3], log [base 9] 81, log... (answered by Earlsdon)

Use your calculator to evaluate each expression a) log(base 2)3000 b) log(base 2)300... (answered by stanbon)