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-   -   Solve x! = 2^x + 8 (http://mymathforum.com/algebra/328330-solve-x-2-x-8-a.html)

 pelluzi February 29th, 2016 12:04 PM

Solve x! = 2^x + 8

Hi there,

I have no idea how to solve this equation:

$\displaystyle x! = 2^x + 8$

Cheers!
Luis P.

 skeeter February 29th, 2016 12:14 PM

x = 4

 pelluzi February 29th, 2016 12:29 PM

Quote:
 Originally Posted by skeeter (Post 524808) x = 4
But how do I prove it?

 skeeter February 29th, 2016 12:33 PM

... solved by observation, x = 4 works.

 mrtwhs February 29th, 2016 12:47 PM

Quote:
 Originally Posted by skeeter (Post 524813) ... solved by observation, x = 4 works.
Since $\displaystyle 120=5!>2^5+8=40$ and factorials grow faster than exponentials, the statement clearly has no solutions for $\displaystyle x \geq 5$

 pelluzi February 29th, 2016 12:54 PM

Quote:
 Originally Posted by skeeter (Post 524813) ... solved by observation, x = 4 works.
I'm sorry, maybe I wasn't clear in my question.

in fact $\displaystyle 4! = 24$, then $\displaystyle 2^4=16$ and $\displaystyle 16+8=24$

Ok, the resolution is intuitive.

But how should I proced in order to prove my result, isolating x, solving the equation etc?

For instance:
(1) $\displaystyle x*(x-1)! = 2^x + 8$

(2) $\displaystyle x = \frac{2^x + 8}{(x-1)!}$

(3) $\displaystyle x = \frac{2^x + 2^3}{(x-1)!}$

(4) $\displaystyle x = \frac{2 * 2^{x-1} + 2^3}{(x-1)!}$

From that point on I dunno how to proced.

 mrtwhs February 29th, 2016 01:12 PM

Quote:
 Originally Posted by pelluzi (Post 524819) I'm sorry, maybe I wasn't clear in my question. in fact $\displaystyle 4! = 24$, then $\displaystyle 2^4=16$ and $\displaystyle 16+8=24$ Ok, the resolution is intuitive. But how should I proced in order to prove my result, isolating x, solving the equation etc?
Some equations cannot be solved in closed form. This is an example of one.

 pelluzi February 29th, 2016 01:23 PM

Quote:
 Originally Posted by mrtwhs (Post 524828) Some equations cannot be solved in closed form. This is an example of one.
Then shall input that equation onto a software that solves by trial an error whenever I face a equation like that? I mean, when it has no intuitive answer.

 Country Boy February 29th, 2016 01:32 PM

1 does not satisfy the equation because $1!= 1\ne 2^1+ 8= 10$.
2 does not satisfy the equation because $\displaystyle 2!= 2\ne 2^2+ 8= 12$.
3 does not satisfy the equation because $\displaystyle 3!= 6\ne 2^3+ 8= 16$.
As skeeter said, 4 does satisfy the equation: $\displaystyle 4!= 24= 2^4+ 8$.

Then Mrtwhs said "Since 120=5!>25+8=40120=5!>25+8=40 and factorials grow faster than exponentials, the statement clearly has no solutions for x≥5."