Verified answer

(x^3-x) ln 5 = ln 1 = 0

x(x^2-1) = 0

x(x+1)(x-1) = 0

x = 0, 1 and -1

5^(x^3 − x) = 1

Taking ln on both sides

=> ln 5^(x^3 − x) = ln 1

=> (x^3 − x) ln 5 = 0 [because ln 1 = 0 and ln constant^variable = variable ln constant]

=> But ln 5 is not = 0

=> (x^3 − x) = 0

=> x^3 = x

=> x^2 = 1

=> x = +1 or x = -1

5^(x^3 -x) = 1

==> 5^(x^3 - x) = 5^0 .... any a^0 = 1

==> (x^3 - x) = 0 ... no need for logs ... and any log, if you use it gets to the same place

==> x(x^2 - 1) = 0

==> x(x+1)(x-1) = 0

==> x = -1 or x = 0 or x = 1

(x³ - x) = 0

x ( x² - 1 ) = 0

x = 0 , x = 1 , x = - 1

Use the fact that ln(a^b) = b ln(a).

In your case, you have a = 5, and b = x^3-x.

5^(x^3-x) =1

(x^3 -x) ln5 = ln1

x^3 -x = 0

x( x^2 -1) =0

x= 0

x^2 =1

x= +_ 1

so x= 0 , 1 , -1 ANSWER