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 ?

  • $\begingroup$ why not $f(3)=27f(1)$ $\endgroup$ – Archis Welankar Dec 22 '15 at 17:01
  • $\begingroup$ How would that help me ?Please elaborate. $\endgroup$ – Mr. Y Dec 22 '15 at 17:02
  • $\begingroup$ do you want a particular value or a range for $a$ $\endgroup$ – Archis Welankar Dec 22 '15 at 17:08
  • $\begingroup$ That's definetly not helping . $\endgroup$ – Mr. Y Dec 22 '15 at 17:10
  • 1
    $\begingroup$ @ArchisWelankar No it doesn't. Consider $f(x)=x^{2}$, $f(1)=f(-1)$. $\endgroup$ – J.Gudal Dec 22 '15 at 17:28

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

  • $\begingroup$ Where does it follow that $1=f(2a)^{2a-1}$ ? $\endgroup$ – Mr. Y Dec 22 '15 at 17:14
  • $\begingroup$ $f(2a)=f(2a)^{2a} \Rightarrow 1=\frac{f(2a)^{2a}}{f(2a)}$ $\endgroup$ – J.Gudal Dec 22 '15 at 17:17

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.

  • $\begingroup$ Thus the second relation let us say that $f(x)=c^x$ is a one-to-one function(so we can take the inverses of both side of the equation) ,right ? $\endgroup$ – Mr. Y Dec 22 '15 at 17:44
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    $\begingroup$ @Mr.Y Essentially so. More specifically, $f(x)=c^x$ is a one-to-one function for all $c\gt 0$ and $c\neq 1$, and the second relation eliminates the possibility that $c=1$ (Note that the original problem says 'for all $x$', which - somewhat sloppily - eliminates the possibility that $c\leq 0$ since the function would not be defined for all values in those cases.) $\endgroup$ – Steven Stadnicki Dec 22 '15 at 19:10

You can't apply f^-1 to both sides since the function isn't 1:1, by your method, then for x=2, a would be +1/sqrt2 or -1/sqrt2. If you try x=3 you also get a different answer. So your method doesn't work at all; horrendous logic.

  • $\begingroup$ Please edit and use MathJax to properly format math expressions. $\endgroup$ – Lee David Chung Lin Jun 9 at 0:14

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