# F(a+b)=f(a)*f(b) and f(a*b)=f(a+b)

1. Mar 16, 2012

### Emilijo

Does sombody know what function has these characteristics:

f(a+b)=f(a)*f(b) and

f(a*b)=f(a+b)

2. Mar 16, 2012

### szynkasz

Re: xyxx

$$f(a+b)=f(a\cdot b)=f(a)\cdot f(b)\\ f(a)=f(a+0)=f(a)\cdot f(0)=f(a\cdot 0)=f(0)\Rightarrow f(0)=1\Rightarrow\fbox{f(a)=1}$$

3. Mar 17, 2012

### lugita15

Re: xyxx

The first statement is satisfied by exponential functions, the second is satisfied by logarithmic functions. It seems like nothing can satisfy both.

4. Mar 17, 2012

### A. Bahat

Re: xyxx

Actually, the second statement is not satisfied by logarithmic functions. You are thinking of f(a*b)=f(a)+f(b), but here it is f(a*b)=f(a+b).

5. Mar 17, 2012

### M Quack

Re: xyxx

f(x)=1 for all x satisfies all conditions. Not a very interesting function...

6. Mar 17, 2012

### SumThePrimes

Re: xyxx

For one that satisfies only the second statement, if f(a+b)=f(a*b), then f(a+0)=f(a*0), so
f(a) is f(0) for any a, so the second statement alone assures us that this function is constant, but there is no way to know what constant with only the fact f(a+b)=f(a*b) .
Now combine your second statement with your first.
Combined with the fact f(a)*f(b)=f(a*b), since f(a)=f(b)=f(a*b)=c for some constant c, as shown by statement 2, c*c=c , there are two solutions for your function when both statements are included, f(x)=1 , and f(x)=0 .