Exponents have a well-known property:

$$x^ax^b = x^{a+b}$$

but

$$x^{a} + x^{b} \neq x^{a+b}$$

Similarly,

$$\log(a) + \log(b) = \log(ab) $$

But

$$\log(a)\log(b) \neq \log(ab)$$

So my question is this:

Is there a function $f$ on $\mathbb{R}$ or some infinite subset of $\mathbb{R}$ with the following properties

$$(1)\quad f(x)f(y) = f(x+y)$$

$$(2)\quad f(x)+f(y) = f(x+y)$$

ie

$$(3)\quad f(x)+f(y) = f(x)f(y)$$It seems that $(2)$ requires the function to be linear…

**Answer**

Your title expresses interest in “a function in which addition and multiplication behave the same way”. That’s condition (3) *alone*. Conditions (1) and (2) are unnecessarily-strong requirements that artificially restrict the possible solutions. Be that as it may …

Let’s invoke condition (3) with three *arbitrary* values, $x$, $y$, $z$.

$$\begin{align}

f(x) + f(y) = f(x)\cdot f(y) \\

f(x) + f(z) = f(x)\cdot f(z)

\end{align}$$

Subtracting, we get

$$f(y) – f(z) = f(x)\cdot(\;f(y)-f(z)\;)\quad\to\quad\left(f(x)-1\right)\cdot\left(f(y)-f(z)\right) = 0$$

For all choices of $x$, $y$, $z$, at least one factor must vanish. We conclude that $f$ must be *some* constant; say, $k$. (The vanishing of the first factor requires specifically that $k=1$, but we’ll go ahead and absorb this into the more-general statement.)

Then condition (3) reduces to

$$k + k = k\cdot k \quad\to\quad k(k-2) = 0$$

so that $k = 0$ or $k = 2$. That is, we have two ways to satisfy condition (3):

$$f(x) \equiv 0 \qquad\text{or}\qquad f(x) \equiv 2$$

Imposing conditions (1) and (2) limits the solutions to just the first.

**Attribution***Source : Link , Question Author : crackpotHouseplant , Answer Author : nneonneo*