Finding the number of real zeros for the equation $11^x + 13^x + 17^x-19^x=0 $ This is a question from a mathematics contest where we are asked about the zeros of the equation $ 11^x + 13^x + 17^x -19^x=0.$
The options are:
1: No real root
2:Only one real root
3: Two real roots
4: More than two real roots.
I want to know a general method to solve these type of questions. I know that if we were given some other function say polynomial, or trigonometric function, etc then we can look into derivative of the function and then by analysing the graph or the maxima and minima we can say about the zeros.
However in this case I am unable to use such things. Any help is highly appreciated.
 A: 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

*

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


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


*$19^x>0$ for all $x\in\mathbb{R}$
we can say that the equation has exactly one real root.
A: 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.
A: 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.
