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 February 29th, 2016, 12:04 PM #1 Newbie   Joined: Feb 2016 From: Brazil Posts: 6 Thanks: 0 Solve x! = 2^x + 8 Hi there, I have no idea how to solve this equation: $\displaystyle x! = 2^x + 8$ Cheers! Luis P. February 29th, 2016, 12:14 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 x = 4 Thanks from pelluzi February 29th, 2016, 12:29 PM   #3
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 Originally Posted by skeeter x = 4
But how do I prove it? February 29th, 2016, 12:33 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 ... solved by observation, x = 4 works. February 29th, 2016, 12:47 PM   #5
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 Originally Posted by skeeter ... 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 $\displaystyle x \geq 5$ February 29th, 2016, 12:54 PM   #6
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 Originally Posted by skeeter ... 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. February 29th, 2016, 01:12 PM   #7
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 Originally Posted by pelluzi 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. February 29th, 2016, 01:23 PM   #8
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 Originally Posted by mrtwhs 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. February 29th, 2016, 01:32 PM #9 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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. February 29th, 2016, 01:49 PM #10 Math Team   Joined: Jul 2011 From: Texas Posts: 2,947 Thanks: 1555 Thanks from pelluzi Tags equation, factorial, solve Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post gen_shao Algebra 12 November 2nd, 2014 06:11 AM

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