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

  1. Mar 16, 2012 #1
    Does sombody know what function has these characteristics:

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

  2. jcsd
  3. Mar 16, 2012 #2
    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}
  4. Mar 17, 2012 #3
    Re: xyxx

    The first statement is satisfied by exponential functions, the second is satisfied by logarithmic functions. It seems like nothing can satisfy both.
  5. Mar 17, 2012 #4
    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).
  6. Mar 17, 2012 #5
    Re: xyxx

    f(x)=1 for all x satisfies all conditions. Not a very interesting function...
  7. Mar 17, 2012 #6
    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 .
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