An Exponentially Hard Logarithm Problem

[PlushChronicles]

What program are you using!??

[boborulllz]

From the second one we substitute the a0 found previously and we get: ln(e^(1/xc))*e((1/xc)*xc)=1, which is (1/xc)*e=1, so finally xc=e, and so if we want to find the value of a0, we find that a0=e^(1/e).

So there it is!

[boborulllz]

Hey mate!

Cool problem, and I think I have found a variation of the solution which doesn't imply guessing. So, let's call the value of x at which the 2 graphs intersect xc, and the value of a for which the 2 graphs intersect at 1 point a0. So, considering that the derivatives of both functions at xc is 1, we can write: 1/(xc*ln(a0))=1, which gives xc*ln(a0)=1, so ln(a0)=1/xc, giving us a0=e^(1/xc).

We also have a second equation: ln(a0)*a0^xc=1.

[jimmayl1]

Very cool problem!

[xtremetvfan]

i dunno about all these maths terms but this vid sure gave me a natural log.

[champion32panel]

Have you done a literature search on this? If so, and you haven't found anything, have you submitted a paper on this?

[AlexPleaseHelpDotCom]

No I haven't. I'm so busy!