Question

Number of real solutions of the equation ${|x-2|}^{x^2-6x+8}=1$ is/are

• A. $2$
• B. $3$
• C. $4$
• D. $1$

Answer 1

The correct option is B $3$

Given $|x-2{|}^{x^2-6x+8}=1$

Now we know that $a^0=1$

Hence $x^2-6x+8=0⟹(x-4)(x-2)=0$

$x=2,4$

When $x=2,|x-2|=0$. This is not possible. Hence $x=2$ is not a solution.

When $x=4$, $|x-2{|}^{x^2-6x+8}=2^0=1$

Another solution is when $|x-2|=1$. Then $x=1,3$

When $x=1$, $|x-2{|}^{x^2-6x+8}=1^3=1$

When $x=3$, $|x-2{|}^{x^2-6x+8}=1^{-1}=1$

Hence solutions are $x=1,x=3,x=4$. Total number of solutions $=3$