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?
 A: If you know the basic properties of exponential functions:


*

*an exponential function of the form $a^x$ ($a>1$) is strictly increasing;

*the sum of two strictly increasing functions, is strictly increasing.

A: It is not restrictive to assume $x\ge y$. If $t=x-y$, we have
$$
3^y(3^t-1)+5^{y}(5^t-1)=0
$$
Since $t\ge0$, also $3^t-1\ge0$ and $5^t-1\ge0$.
A: $f(x)$ is injective means $f(x) = f(y) \iff x = y$.
So if you can show $x \ne y \implies f(x) \ne f(y)$ you are done.
Let $x < y$.  Then $y = x + k$ for some $k> 0$.  So $3^y = 3^{x+k} = 3^x\cdot 3^k$.  And $3^k > 1$ so $3^x3^k > 3^x$.  So $3^y > 3^x$.  
By the same reasoning $5^y > 5^x$ and therefore $3^y + 5^y > 3^x + 5^y$.
That's it.
A: Compute the derivative of $f(x)=\exp(x\ln(3))+\exp(x\ln(5)).$ Its derivative is $\ln(3)\exp(x\ln(3))+\ln(5)\exp(x\ln(5))>0.$
A: 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 $3^x+5^x=3^y+5^y\Rightarrow x=y\space\space\space \blacksquare$
