Desmos won't help you this time!

Either $(anything) ^0$ , or $(1)^{anything}$ , or $(-1)^{even integer}$

Solve the quadratics for these values, and you'll get the answers as $2,3,4,5,16$

Let $f(x) = g(x)^{h(x)}$ , where $g(x) = x^2 - 7x+11$ and $h(x) = x^2-19x+48$ . We note that $f(x) = 1$ , when

$\begin{cases} f(x) = 1^{h(x)} & \text{i.e. }g(x) = 1 \text{ for all real } h(x) \\ f(x) = (-1)^{2n} & \text{i.e. }g(x) = -1 \text{ and }h(x) = \text{ even integer} \\ f(x) = g(x)^0 & \text{i.e. }h(x) = 0 \text{ for all real } g(x) \text{ except 0} \end{cases}$

Case $f(x) = 1^{h(x)}$ :

$\begin{aligned} \implies g(x) & = 1 \\ x^2-7x+11 & = 1 \\ x^2-7x+10 & = 0 \\ (x-2)(x-5) & = 0 \end{aligned}$

$\implies x = \begin{cases} 2 & \implies h(2) = 14 & \implies f(2) = 1 \\ 5 & \implies h(5) = -22 & \implies f(5) = 1 \end{cases}$

$\begin{aligned} \implies g(x) & = -1 \\ x^2-7x+11 & = -1 \\ x^2-7x+12 & = 0 \\ (x-3)(x-4) & = 0 \end{aligned}$

$\implies x = \begin{cases} 3 & \implies h(3) = 0 & \implies f(3) = 1 \\ 4 & \implies h(4) = -12 & \implies f(4) = 1 \end{cases}$

Case $f(x) = g(x)^0$ :

$\begin{aligned} \implies h(x) & = 0 \\ x^2-19x+48 & = 0 \\ (x-3)(x-16) & = 0 \end{aligned}$

$\implies x = \begin{cases} 3 & \implies g(3) = -1 & \implies f(3) = 1 \\ 16 & \implies g(16) = 155 & \implies f(16) = 1 \end{cases}$

Therefore, there are $\boxed 5$ unique solutions of $x=2,3,4,5,16$ .

As noted by @Parth Sankhe , if $a^b=1$ for real $a$ and $b$ , then one of three cases applies:

• $a=1$ -- then $x^2-7x+11=1\implies (x-2)(x-5)=0$ and $x=2$ or $x=5$ .
• $b=0$ -- then $x^2 -19x + 48 = 0 \implies (x-3)(x-16)=0$ and $x=3$ or $x=3$ (note that neither of these are roots of $x^2 - 7x + 11$ )
• $a=-1$ and $b$ is even -- then $x^2-7x+11=-1 \implies (x-3)(x-4) = 0$ and $x=3$ or $x=4$ . We already know that $x=3$ is a solution. Substituting $x=4$ yields $(-1)^{68}$ a solution.

Thus, there are $\boxed{5}$ solutions.