Solving An Interesting Exponential Equation

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[yurenchu]

(x - 3)⁶ = (3 - x)ˣ

Upon inspection, clearly x = 3 is a solution, because then both sides become 0 .
Now for other solutions, we can assume x ≠ 3 and therefore we can safely divide both sides by (3 - x)⁶ :

1 = (3 - x)ˣ⁻⁶

The solutions to this equation are given by:
x - 6 = 0 ==> x = 6
(3 - x) = 1 ==> x = 2
(3 - x) = -1 AND (x-6) is even ==> x = 4

Plugging these values for x into the original equation confirms that these are indeed solutions.

So the complete set of solutions is:
x = 2 OR x = 3 OR x = 4 OR x = 6

[RyanLewis-Johnson-wqxs]

(x-3)^6=(3-x)^x x=2 x=3 x=4 x=6 final answer

[Don-Ensley]

problem
(x-3)⁶ =(3-x)ˣ

Replace x-3 with (-1)(3-x).
(3-x)⁶ =(3-x)ˣ

Let u = 3-x
First solution is when u = 0 is when x=3.
We have
(0)⁶ =(0)³

u⁶ = u ³⁻ᵘ
u⁻ ³ = u ᵘ
u ⁽³⁺ᵘ⁾ = 1

3 possibilities of base raised to power:

u = 1 because 1 raised to any power = 1
3+u = 0, because anything to the 0 is 1
u = -3
u is negative and 3+u is even.
u = -1 works

All the u solutions are
u = -3
u = -1
u = 0
u = 1

Back substitute
x = 3-u

x = 6
x = 4
x = 3
x = 2

answer

x ∈ { 2, 3, 4, 6 }

[АндрейПергаев-зн]

Автор не совсем прав
Если (-1)^а=1 если а чётное число, что не было сказано
Соглашусь что ответ верный, но иногда бывает по другому