Question

The number of values of $x$ satisfying the equation $(x^2+7x+11{)}^{(x^2-4x-21)}=1$ is

• A. $4$
• B. $6$
• C. $5$
• D. $2$

Answer 1

The correct option is A $4$

$(x^2+7x+11{)}^{(x^2-4x-21)}=1$
Let $a=x^2+7x+11$ and $b=x^2-4x-21$

Case $1:$ $a^b=1$ when $a=1$
$x^2+7x+11=1\Rightarrow x^2+7x+10=0\Rightarrow (x+2)(x+5)=0\Rightarrow x=-5,-2$

Case $2:$ $a^b=1$ when $b=0$ and $a\ne 0$
$x^2-4x-21=0\Rightarrow (x+3)(x-7)=0\Rightarrow x=-3,7$

Case $3:$ $a^b=1$ when $a=-1$ and $b$ is an even number.
$x^2+7x+11=-1\Rightarrow x^2+7x+12=0\Rightarrow (x+3)(x+4)=0\Rightarrow x=-3,-4$
$x=-3\Rightarrow b=0$ which is even.
$x=-4\Rightarrow b=11$ which is odd.
So, $x=-4$ doesn't satisfy the assumed conditions.

So, the solutions are $-5,-2,-3,7$