My coworker and I are math idiots. Algebra help.

[ggavinmoss]

x^(1/n) = y

Solve for n.

Don't tell me to do my own homework, it's for work and I'm tired.
Your help is appreciated.

-geoff

 

[conjur]

Do your own homework

 

[ggavinmoss]

Originally posted by: conjur
Do your own homework

You bastard.

We were thinking this:

x^(1/n) = y
x = y^n
log(x) = log(y^n)
log(x) = n * log(y)
n = log(x) / log(y)

?

-geoff

 

[BigPoppa]

Take the natural log of both sides, which will bring the power to the front. (1/n)ln(x)=ln(y). Divide by ln(x), (1/n) = ln(y)/ln(x). Invert. n = ln(x)/ln(y)

 

[ggavinmoss]

Originally posted by: BigPoppa
Take the natural log of both sides, which will bring the power to the front. (1/n)ln(x)=ln(y). Divide by ln(x), (1/n) = ln(y)/ln(x). Invert. n = ln(x)/ln(y)

Technically, can't it be log base anything?

-geoff

 

[InverseOfNeo]

Originally posted by: BigPoppa
Take the natural log of both sides, which will bring the power to the front. (1/n)ln(x)=ln(y). Divide by ln(x), (1/n) = ln(y)/ln(x). Invert. n = ln(x)/ln(y)

WOW
/me gazes in awe of a persons knowledge of logs

 

[BigPoppa]

Yes, you can use log base anything Gavin. I just have always used natural log.