# Continuous function $g$ satisfying $g(x + y) = 5g(x)g(y)$

Let $g$ be a continuous function with $g(1) = 1$ such that $$g(x + y) = 5g(x)g(y)$$ for all $x$, $y$. Find $g(x)$.

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Please explain the solution as much as you can – Argha Jul 17 '12 at 3:33
Here's a hint: Work out what it is for integers first - you will spot a formula for $g$. Then prove that this also has to hold for all rationals. By continuity, you then have the unique function $g$ satisfying your constraints. – us2012 Jul 17 '12 at 3:44

setting $y=1$ gets
$$g(x+1)=5g(x)g(1)=5g(x)$$
So every time you increase the argument by $1$, you multiply by $5$. Can you see what function has this property? and how to prove that the solution is unique? You probably won't get complete answers until you post some of your work, so we know what specifically to help you with.
A small beginning: Let $g(t)=\frac{h(t)}{5}$. Then our functional equation can be rewritten as $$\frac{h(x+y)}{5}=5\frac{h(x)}{5}\frac{h(y)}{5},$$ or equivalently $$h(x+y)=h(x)h(y).$$ This one is a standard functional equation that has been discussed on this site and elsewhere.