Simple calculus question

[DisgruntledVirus]

So I am doing some calculus homework and am questioning my answer. The problem is:

Express y as a function of x. 3ln y=1/3ln(2x) - ln ((x2-1)/(x^4/3)) + ln C

I got the answer to be y = cubert((C*cubert(2x)*(x2-1))/cubert(x^4)) but for some reason I am second guessing myself on it and I dont know why.

 

[3NF]

y(x) = (2^(1/9) C^(1/3) x^(5/9))/(-1 + 2 x)^(1/3)

 

[Saint Michael]

y = e^(1/9ln(2x) - 1/3ln ((x2-1)/(x^4/3)) + 1/3ln C)

Edit: The 1/3 in the beginning is a little confusing.

Edit 2: And what the hell is cubert?

Edit 3: Oh, cube root.

 

[hypn0tik]

ln(A) - ln(B) = ln(A/B).

ln(A) + ln(B) = ln(A*B)

ln(C^D) = D * ln(C)

ln(AAA) = ln(BBB) --> AAA = BBB

Use those rules of logarithms and you should be set.

 

[Leros]

Originally posted by: hypn0tik
ln(A) - ln(B) = ln(A/B).

ln(A) + ln(B) = ln(A*B)

ln(C^D) = D * ln(C)

ln(AAA) = ln(BBB) --> AAA = BBB

Use those rules of logarithms and you should be set.

The properties of logarithms are simply amazing and so useful in manipulating equations.

 

[Kelemvor]

Wow. i haven't played CuBert in years... Isn't it QBert though?

 

[hypn0tik]

Originally posted by: Leros

Originally posted by: hypn0tik
ln(A) - ln(B) = ln(A/B).

ln(A) + ln(B) = ln(A*B)

ln(C^D) = D * ln(C)

ln(AAA) = ln(BBB) --> AAA = BBB

Use those rules of logarithms and you should be set.

The properties of logarithms are simply amazing and so useful in manipulating equations.

Yep. Makes the differentiation of messy functions easy as well!