Matematické Fórum

stuart clark · 10. 08. 2014 06:31

(1) Find the integer solutions of the equation $3^x+x^4 = 5^x$

(2) The number of real solution of $2^{2^{x}} = 4x^2$

Lukáš Ba-mat-fyz

↑ stuart clark:

The only advice that I can give you is to substitute in (2) $2x=t$ a then you get equation where you can see 2 solutions, but there could be more.

For (1) you can try few numbers and you will find two nice solutions but now I do not know procedure.

Brzls · 11. 08. 2014 11:29

2) Easy to check that on (0, infinity) are both functions increasing and convex, so there can be one or two solutions. We are able to find two solutions (1,2), so there are two solutions on (0, infinity)
On (-infinity,0) 4x^2 is strictly decreasing with range of values (0, infinity) and function 2^2^x is increasing with range of value (0,4) so there is exactly one real solution.
So $2^{2^{x}} = 4x^2$ has 3 solutions.

1) Both functions are also increasing and convex, so there can be only one or two solutions on (0,infinity). First is x=2, and one can easily check, that the second lie between 2 and 3 (with suitable numericalmethod for example). Obviously there is no solution wchich lie in (-infinity,0)
Integer solutions are 0 and 2

Xellos · 11. 08. 2014 18:46

One more nice observation for 1): taking the equation modulo 8, we can see that for $x$ odd, $x^4\equiv x^2\equiv 1 \mod 8$ and $3^x \equiv -(5^x) \equiv 3 \mod 8$ , so the equation has no solution for $x$ odd.

Matematické Fórum

#1 10. 08. 2014 06:31

Exponential Equation.

#2 11. 08. 2014 09:11 — Editoval Lukáš Ba-mat-fyz (11. 08. 2014 09:16)

Re: Exponential Equation.

#3 11. 08. 2014 11:29

Re: Exponential Equation.

#4 11. 08. 2014 18:46

Re: Exponential Equation.

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