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Let $f(x+1)=f(x)\cdot f(y)$, for all $x,y\in\mathbb{R}$. If $f(1)=8$ then find $f\left(\frac{2}{3}\right)$.

I'm trying to find it as below.

\begin{align} f(1)=f(0+1)=f(0)\cdot f(y)=8\\ f\left(\frac{2}{3}\right) = f\left(-\frac{1}{3}+1\right)=f\left(-\frac{1}{3}\right)\cdot f(y). \end{align}

Since I can't find $f(y), f(0), f\left(-\frac{1}{3}\right)$, now I confused how to find $f\left(\frac{2}{3}\right)$. Any hint to solve that?

Ongky Denny Wijaya
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    This functional equation seems to have atypo - because what would happen if I have two different values of $y$ for the same $x$? $f(y_2) = f(y_1)$ for all $y$ means the function is constant and hence 8 everywhere – Dhanvi Sreenivasan Jan 29 '21 at 07:24
  • @Chrystomath: the question you linked is a different one. Notice $f(x+1)$ instead of $f(x+y)$. – Prasun Biswas Jan 29 '21 at 07:33

2 Answers2

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With $y=1$, the functional equation becomes $$f(x+1)=8f(x)$$ for all $x$.In particular, $f(2)=64$. Then with $x=1$, $y=2$, the functional equation becomes $$ 64=8\cdot64$$ contradcition. No such $f$ exists.

Hagen von Eitzen
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$f(1)=f(0)f(y)$ for all $y$. So $f$ is a constant function but the only contsant functions satifying the hypothesis are $0$ and $1$ so we cannot have $f(1)=8$.

Kavi Rama Murthy
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