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.
 A: Rewrite the equation as
$$f(x)=\left({2\over5}\right)^x + \left({3\over5}\right)^x + \left({4\over5}\right)^x = 1$$
Since the fractions being exponentiated are all less than $1$, $f$ is a strictly decreasing function of $x$, hence can take the value $1$ at most once.  It's clear that $f(0)=3\gt1$ and $f(x)\rightarrow0$ as $x\rightarrow\infty$, so $f(x)=1$ has exactly one solution.  If you need to narrow it down, note that $f(2)=(4+9+16)/25=29/25\gt1$ while $f(3)=(8+27+64)/125=89/125\lt1$, so the single solution lies in the range $2\lt x\lt3$.  
A: Alternate form: $2^x+2^{2x}+3^x=5^x$
plotting graph using wolfram:
 $x\approx2.37329$
So there's just one solution.
A: 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. 
A: $2^x$+$3^x$+$4^x$=$5^x$


*

*now x cannot be odd because 5|$2^x$+$3^x$ for all odd x.then 5|$4^x$.this cannot be possible.

*so we consider x=2a.now $4^a$$\equiv$$\pm$1(mod 5) and $9^a$$\equiv$$\pm$1(mod 5) and $16^a$$\equiv$1(mod 5)

*so $2^x$+$3^x$+$4^x$$\equiv$$\pm$1$\pm$1+1(mod 5)

*so we have 5|$\pm$1$\pm$1+1.so this cannot be possible

*so x is not even also.so the equation have no solution

