$f(a+b) = f(a) \cdot f(b)$

Question

So I'm working on a question:

$f(1) \cdot f(2) = 8$

What is $f(1) \cdot f(3)$?

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I've figured out that $f(3) = 8$ but I can't seem to play with the fuctions so that I get $f(4)$, which should be $f(1) \cdot f (3)$.

I'm sure it's some really simple arithmetic thing I'm missing.

is $f(x)$ is a real function, it must be if the form $f(x)=k^x$ — WW1, May 10 '18 at 21:15
@WW1 Which is false. The formula only holds for rational $x$. — freakish, May 10 '18 at 21:15
You do have an exponential function $a^x$ and the value of $a$ is almost done. — Piquito, May 10 '18 at 21:21

freakish · Accepted Answer · 2018-05-10T21:20:44.777

5

Note that if $n\in\mathbb{N}$ then

$$f(n)=f(1+\cdots+1)=f(1)\cdots f(1)=f(1)^n$$

and thus by your assumption

$$8=f(1)f(2)=f(1)f(1)^2=f(1)^3$$

and therefore $f(1)=2$ (assuming we are working over $\mathbb{R}$) and now we can apply the original formula again to obtain:

$$f(2)=4$$ $$f(3)=8$$ $$f(4)=16$$ $$\cdots$$ $$f(n)=2^n$$

Note that this can be extended to all rationals, i.e. if $q\in\mathbb{Q}$ then $f(q)=f(1)^q$. However this is no longer true for all reals (unless additional conditions are assumed, e.g. continuity of $f$).

edited May 10 '18 at 21:20

answered May 10 '18 at 21:15

freakish

42,851

That's it :D I knew it was something simple like that. Thank you – Eli May 10 '18 at 21:17

$f(a+b) = f(a) \cdot f(b)$

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