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Challenge (calculus)

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For what value of b is the line y = 10x tangent to the curve y = e^(bx)?
@hartnn, what are your thoughts about it?

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Other answers:

10 = be^(bx) and maybe to find point of intersection, 10x = e^(bx) still thinking further steps...
Medal is awaiting .. . :)
Haha, screw the medal :P
Oh wait I wrote the wrong thing down. I meant to write down what was in the attachment... one sec.
For what value of b is the line y = 5x tangent to the curve y = e^(bx)?
Same logic though.
yeah, idd, phew :P
At the required point, e^(bx) has a slope of 10 => be^(bx) = 10. We can solve for b using newton's method.
5 =be^(bx) 5x = e^bx ---> bx =1 ---->5x = e^1 x= e/5 b = 5/e
You have demonstrated your ability. Medal earned!
lol thanks :)
Although I like the idea of using Newton's method.. Sometimes it doesn't work though.
I agree. There are known strict conditions of convergence. We need to get a close starting point. In any case, we can solve for b=5/e here, but still you need to calculate x one way or another.
Good question btw, I like. Now back to drinking beer.

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