If,for all $x$,$f(x)=f(2a)^x$ and $f(x+2)=27f(x)$,then find $a$

Question

I am finding this problem confusing :

If,for all $x$,$f(x)=f(2a)^x$ and $f(x+2)=27f(x)$,then find $a$.

When $x=1$ I have that $f(1)=f(2a)$ using the first identity.

Then when $x=2a$ I have by the second identity that $f(2a+2)=27f(2a)$,after that I simple stare at the problem without having a clue of how to proceed.

What's the trick the problem is calling for ?

I've thought of finding the inverse of the function $f(x)$ but It's not really clear to me how to apply this idea as I don't have linear functions .

Can you guys give me a hint ?

Answer 1

Note that:

$f(2a)=f(2a)^{2a} \Rightarrow 1=f(2a)^{2a-1}$

Hence either $f(2a)=1$ or $2a-1=0$.

But if $f(2a)=1$, then,

$f(x)=1^{x}$ for all $x$, but $f(2)=27$ and so this is false.

Consequently $a=\frac{1}{2}$.

Answer 2

In fact, you barely even need the second relation. Hint: first set $f(2a)=c$. Now you know that $f(x)$ is an exponential function, $f(x)=c^x$ (and the second relation implies that $c\neq 1$ — this is all it does; 27 could as easily be 0.27 or $10^{27}$ and it wouldn't change the answer); this function is one-to-one over its domain, and you've already figured out that $f(1)=f(2a)$. Apply $f^{-1}$ to both sides.