Finding the number of real zeros for the equation $11^x + 13^x + 17^x-19^x=0 $

THe derivative method is, probably the best one. But here, you may begin with some algebraic trick

REcall that $$11^x + 13^x + 17^x -19^x=0 \iff \left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1=0.$$ Define now $F(x):=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$ wich is continuous and differentiable on $\Bbb R$.
$F(0)=2>0$ and $\lim_{x\to+\infty}F(x)=-1<0$ (so, there is $M\in\Bbb R$ such that $F(M)<0$). An application of Bolzano's Theorem leads to the existence of, at least, one real (positive, in fact) solution.

But, $F'(x)=\left(\frac{11}{19}\right)^x\ln(11/19)+\left(\frac{13}{19}\right)^x\ln(13/19)+\left(\frac{17}{19}\right)^x\ln(17/19)<0$ for all $x\in\Bbb R$, so the solution is unique.

We can rewrite the equation: $$11^x+13^x+17^x-19^x=0\implies 19^x\left[\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1\right]=0$$

and note that $$f(x)=\left(\frac{11}{19}\right)^x+\left(\frac{13}{19}\right)^x+\left(\frac{17}{19}\right)^x-1$$ is a continuous, strictly decreasing function. With

1. $\lim_{x\to-\infty}f(x)=\infty$

2. $\lim_{x\to\infty}f(x)=-1$

3. $19^x>0$ for all $x\in\mathbb{R}$

we can say that the equation has exactly one real root.

Let $$f(x)=(11/19)^x+(13/19)^x+(17/19)^x-1$$ $$\implies f'(x)=(11/19)^x \ln (11/19)+ (13/19)^x \ln (13/19)+(17/19)^x \ln (17/19)<0$$ Hence $f(x)$ is monotonically decreasing so $f(x)=0$ will have atmost one real root and $f(0)=2 >0$ and $f(\infty)=-1<0.$ So by IVT, $f(x)=0$ has one real root. Combining both statements $f(x)=0$, has exactly one real root.