Find number of solutions of $2^x$+$3^x$+$4^x$=$5^x$ Find number of solutions of
$$2^x+3^x+4^x=5^x$$
I tried using graphs but don't know how to draw graph of L.H.S.

You can prove it with the intermediate value theorem.
Suppose $f(x) = 2^x + 3^x + 4^x$ and $g(x) = 5^x$. Both are continuous and differentiable for all $x \in \mathbb{R}$. We see that $f(0) = 1 < 3 = g(0)$ and $f(3) = 99 < 125 = g(3)$. So there is at least one solution to $f(x) = g(x)$. Let us call the smallest solution to the equation $z$. We see that 
$$
f^{(k)}(x) = (\ln 2)^k \cdot 2^x + (\ln 3)^k \cdot 3^x + (\ln 4)^k \cdot 4^x
$$
and
$$
g^{(k)}(x) = (\ln 5)^k \cdot 5^x.
$$
Now we see that
\begin{align*}
f^{(k)}(z) &< (\ln 5)^k \cdot ( 2^z + 3^z + 4^z) \\
&= (\ln 5)^k \cdot 5^z \\
&= g^{(k)}(z),
\end{align*}
so we know the curves cross rather than being tangential. For values of $x \geq z$, we see that $f(x) < g(x)$ and $f'(x) < g'(x)$, so there cannot be any other solutions to $f(x) = g(x)$. Therefore, there is only one solution. 
