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February 29th, 2016, 12:04 PM   #1
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Solve x! = 2^x + 8

Hi there,


I have no idea how to solve this equation:

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



Cheers!
Luis P.
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February 29th, 2016, 12:14 PM   #2
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x = 4
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February 29th, 2016, 12:29 PM   #3
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Quote:
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x = 4
But how do I prove it?
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February 29th, 2016, 12:33 PM   #4
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... solved by observation, x = 4 works.
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February 29th, 2016, 12:47 PM   #5
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Quote:
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... 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 $\displaystyle x \geq 5$
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February 29th, 2016, 12:54 PM   #6
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Quote:
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... 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.
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February 29th, 2016, 01:12 PM   #7
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Quote:
Originally Posted by pelluzi View Post
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.
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February 29th, 2016, 01:23 PM   #8
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Quote:
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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.
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February 29th, 2016, 01:32 PM   #9
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1 does not satisfy the equation because .
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."

That answers your question.
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February 29th, 2016, 01:49 PM   #10
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