Question

solve this equation for x, (x^2-5x+5)^(x^2-2x-48) = 1

Answer 1

Given that

$(x^2-5x+5)^{x^2-2x-48} = 1$

This can be written as

$(x^2-5x+5)^{x^2-2x-48} = (x^2-5x+5)^{0}$

That implies,

$x^2-2x-48 = 0$

solving this equation using factorisation method,

(x-8)(x+6) = 0

x = 8 or x = -6

Answer 2

x²-5x+5^x²-2x-48=1
(x²-5x+5)^(x²-2x-48)=(x²-5x+5)^0
because anything power zero is 1
x²-2x-48=0
by factorization
x²-8x+6x-48=0
x(x-8)+6(x-8)=0
(x+6)(x-8)=0
by using the principle of zero products
x+6=0   or    x-8=0
x=-6      or       x=8
hence the required values of x are -6,8