2^n + n = 40, find n. Please show work!!
-
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.
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.
-
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…
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…
-
I used PARI to get n = 5.125, approximately.
A problem like this is best done graphically.
A problem like this is best done graphically.