Challenging Problem?!
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Challenging Problem?!

[From: ] [author: ] [Date: 11-08-09] [Hit: ]
Another way to try it,or 0.But since you know that n is just a little larger than 5,Then n = ( log 35)/.and youll be very close to the answer above.[ESP5.......
2^n + n = 40, find n. Please show work!!

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The answer won't be a whole number because 2^5 = 32 and 2^6 = 64, so n has to be very close to 5.
By trial and error, you'll find that n is approximately 5.125

Another way to try it, is to write the equation as 2^n = 40 -n
Take the log of both sides: n log 2 = log (40 -n)
or 0.301n = log (40-n)
But since you know that n is just a little larger than 5, you can approximate the right side of the equation by finding log 35
Then n = ( log 35)/.301
and you'll be very close to the answer above.

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2^6 = 64 so 6 is too big

2^5 = 32 and 32 + 5 = 37 so 5 is too small

so n must be somewhere between 5 and 6

but very close to 5

2^n + n - 40 = 0

[ESP 5.02 keeps popping up for a try, but I don't know....]

How do you compute 2^5.02 ????

(2)^502/100 = 100th root of 2^502 ???

No one ever showed the bear how to do this.

Maybe something here will help ---> you <---
http://www.mathsisfun.com/algebra/expone…

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I used PARI to get n = 5.125, approximately.
A problem like this is best done graphically.
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