Showing $3^x+5^x$ is injective How can we show that
$$f(x)=3^x+5^x$$
is injective?
I know that $f$ is strictly increasing in $R$ so $f(x) = k, k \in R$ has only one root.
But I was looking for a proper proof, something I tried was:
Let $x,y \in \mathbb R$. Then for $f$ to be injection, $$3^{x}+5^{x} = 3^{y}+5^{y}$$ must imply that $x=y$.
Now how do we proceed?

Suppose $3^x+5^x=3^y+5^y.$
Now $$3^x+5^x=3^y+5^y\Rightarrow 3^x-3^y=-(5^x-5^y)$$
Case 1: $x\geq y$, then $5^x-5^y\geq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\leq 0$. But  $3^x-3^y\geq 0$. Which lead to  $3^x=3^y\Rightarrow x=y$
Case 2: $x\leq y$, then $5^x-5^y\leq 0\Rightarrow 3^x-3^y=-(5^x-5^y)\geq 0$. But  $3^x-3^y\leq 0$. Which lead to  $3^x=3^y\Rightarrow x=y$
Hence $3^x+5^x=3^y+5^y\Rightarrow x=y\space\space\space \blacksquare$
