Then tried taking the derivative of each side, but it didn't work.
You could do a Taylor expansion on the ln (use base 2!) to get ln(x) in polynomial form and then add in xln2 into that and reverse the Taylor. That might work.
It can also be recast as e^(ln2*x) = 5x, and e is of course defined as the series x^n/n! where n is summed from 0 to infinity. Numerics win here.
That's exactly where I got -- but I'm just too lazy to work out the Taylor series. Shudder....Taylor series. It's not hard, just too much to crunch and write out. I remembered it enough for the AP Calculus BC exam -- only came up on one friggin question.
Comments
Originally posted by MajorMatt
There are two answers, which I forgot.
Anyhow plotting is breaking the rules of this exercise.
I was looking a plot when I noodled out there were two answers.
Originally posted by Rick1138
It's pretty damn difficult.
I got it into the form:
ln 5x = x ln2
Then tried taking the derivative of each side, but it didn't work.
You could do a Taylor expansion on the ln (use base 2!) to get ln(x) in polynomial form and then add in xln2 into that and reverse the Taylor. That might work.
http://mathworld.wolfram.com/LambertsW-Function.html
It can also be recast as e^(ln2*x) = 5x, and e is of course defined as the series x^n/n! where n is summed from 0 to infinity. Numerics win here.
It can also be recast as e^(ln2*x) = 5x, and e is of course defined as the series x^n/n! where n is summed from 0 to infinity. Numerics win here.
That's exactly where I got -- but I'm just too lazy to work out the Taylor series. Shudder....Taylor series. It's not hard, just too much to crunch and write out. I remembered it enough for the AP Calculus BC exam -- only came up on one friggin question.
Bah.