Question

Find the real roots of the equation $7^{{\log }_7(x^2-4x+5)}=(x-1)$.

Answer 1

$7^{{\log }_7(x^2-4x+5)}=(x-1)$

or, $x^2-4x+5=x-1$ [Since $a^{{\log }_{a}x}=x$ for $a>0,x>0$]

or, $x^2-5x+6=0$

or, $(x-2)(x-3)=0$

or, $x=2,3$.

So the real roots are $2,3$.