What is the solution sets for $(5x-2)^{x-5}=(5x-2)^{2x+1}$?

Question

Here is my attempt:

For every exponent equation like this, we can use 4 equations to find its solution sets:

The solution sets of $$(a(x))^{f(x)}=(a(x))^{g(x)}$$ is satisfy:

1. f(x) = g(x)

2. a(x) = 1

3. a(x) = -1, with the criteria f(x) and g(x) are both odd or even

4. a(x) = 0, with the criteria f(x) > 0 and g(x) > 0

For number 1, we got:

x-5 = 2x+1

x = -6

For number 2, we got:

5x-2 = 1

x = $3/5$

For number 3, we got:

5x-2 = -1

x = $1/5$

Criteria: f(x) and g(x) both odd or even.

$$f(1/5)= -24/5$$ and $$g(1/5) = 7/5$$

Or we can say $f(1/5)= -4.8$ and $g(1/5)=1.4$

f(x) and g(x) both even

For number 4, we got:

a(x) = 0

5x-2= 0

$x = 2/5$

Criteria f(x) and g(x) must be larger than 0

f(x)=0

$x =5 --> f(x) > 0$

g(x) = 0

$x =-(1/2)--> g(x) < 0$

Because g(x) < 0, then $x=2/5$ isn't a solution.

So, the solution sets are ${(-6, 1/5, 3/5)}$

Is my solution true?

Now, I have another question, when I'm trying to recheck my solution(x=1/5), I've got the equality:

$$(-1)^{-4.8}=(-1)^{1.4}$$

How can I verify the value of $$(-1)^{-4.8}$$ and the value of $$(-1)^{1.4}$$?

Thanks

Answer 1

Hint: ${-1} = e^{i(1+2n)\pi}$ for any arbitrary $n\in\Bbb Z$

If any such $n\in\Bbb Z$ exists that $-4.8 = 1.4(1+2n)$ then the terms are equal. Otherwise they are not.