Solve x! = 2^x + 8 Hi there, I have no idea how to solve this equation: $\displaystyle x! = 2^x + 8$ Cheers! Luis P. 
x = 4 
Quote:

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

Quote:
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) $\displaystyle x*(x1)! = 2^x + 8$ (2) $\displaystyle x = \frac{2^x + 8}{(x1)!}$ (3) $\displaystyle x = \frac{2^x + 2^3}{(x1)!}$ (4) $\displaystyle x = \frac{2 * 2^{x1} + 2^3}{(x1)!}$ From that point on I dunno how to proced. 
Quote:

Quote:

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. 

All times are GMT 8. The time now is 06:52 AM. 
Copyright © 2019 My Math Forum. All rights reserved.