Find the number of solutions of $$2^x+3^x+4^x-5^x=0$$
Answer is given $1$. I tried to take the derivative of the function $$f(x)=2^x+3^x+4^x-5^x\\f'(x)=\ln2 (2^x)+\ln3(3^x)+\ln4(4^x)-\ln5(5^x)$$ but I could not make the conclusion wheather it is $\gt0$ or, $\lt0$. Projecting $f'(x)$ in desmos it is showing both +ve for certain values of $x$ and -ve for other values. So function is not monotonic. I have also projected $f(x)$ in desmos; found $f(x)$ to be first incresing and then decreasing. But how can this be solved without using graphical calculator?
Consider the equation $$\begin{align} 2^x+3^x +4^x &=5^x\\ \left({2\over 5}\right)^x+\left({3\over 5}\right)^x+\left({4\over 5}\right)^x &=1 \tag 1 \end{align}$$
Now consider the function $$\begin{align} g(x) &= \left({2\over 5}\right)^x+\left({3\over 5}\right)^x+\left({4\over 5}\right)^x \\ g'(x) &= \ln\left({2\over 5}\right)\left({2\over 5}\right)^x+\ln\left({3\over 5}\right)\left({3\over 5}\right)^x+\ln\left({4\over 5}\right)\left({4\over 5}\right)^x \end{align}$$ Now all $\ln\left({2\over 5}\right),\ln\left({3\over 5}\right),\ln\left({4\over 5}\right)$ terms are $-ve$ and $k^x;\;\forall k\in \mathbb R$ terms are$+ve$ so $g'(x)\lt0$ hence decreasing function
also for $\begin{cases}x \to -\infty & g(x)\to \infty\\x\to\infty & g(x)\to 0 \end{cases}$
so it will cut the line $y=k ,\; k\in \mathbb R^+$ only once.