1. Just for fun... It was a bonus question on a calc test and I liked it. Maybe those of you for whom it's likely very easy could hold off for a little while and see how other people do with it.

Find the value of b for which 1+e<sup>b</sup>+e<sup>2b</sup>+e<sup>3b</sup>+...=9

2.

3. An excellent question indeed! I like how it turns what would be a fairly routine problem in, say, second semester calculus on its head.

4. Would anyone like to give it a try?

5. I’ll post the answer and leave the working to others who are still interested in giving it a try. 8)

b = ln(8*⁄*9) = −ln(1.125)

6. Crikey this took me a lot of steps. There must be an easier way than what I did.

I refuse to write out my solution without latex, but here is the steps in words:

1. Integrate the original equation,
2. notice the result has the form ln(1+e^b) + stuff,
3. differentiate that,
4. notice a geometric series in the result,
5. here you get "junk = 10",
6. massage "junk" by factoring, canceling, expanding, etc.,
7. you should end up with (e^b+1)(-8+9e^b)=0,
8. BAM! solve for b.

So my solution involved integration, differentiation, use of the series ln(1+x), use of a geometric series, expanding, and factoring.

Cheers,
william

7. Well, actually, there is no need for integration or differentiation. The whole thing is an infinite geometric series.

8. Originally Posted by JaneBennet
Well, actually, there is no need for integration or differentiation. The whole thing is an infinite geometric series.

Shiiiiit! I'll be damned. That makes it a piece of cake! A 10-second problem.

Thanks,
william

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