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Today I've encountered a question like The following;

If function $f$ satisfies $f(xy)=f(x)f(y)$ and $f(81)=3$ then find The value of $f(2)$? What baffles me about this question is that I have to find The equation of the function in order to find $f(2)$ because $2$ is not a divisor of $81$ , using The property I found out that $f(3)=\sqrt[4]{3}$ and wondered if the function could be $f(x)=\sqrt[4]{x}$ (it satisfies the equation up there) but I do not know whether f gives $2$ a value like this or not. And there are many other functions that can be found.

So The question is how can İ get myself out of this ugly situation, and how can I find other $f$ functions that satisfy the constraints? What I am asking is not to prove that these functions are in type of $x^n$ I am trying to get what $f(2)$ is and see also whether this question is deficient ör cannot ve solved with The given details.

Thank you:)


marked as duplicate by GAVD, kingW3, José Carlos Santos, Namaste algebra-precalculus Sep 23 '17 at 0:08

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  • 2
    $\begingroup$ Is $f$ mapping from $\mathbb{R},\mathbb{Q}$, or $\mathbb{Z}$? and is that the only assumption on $f$/ $\endgroup$ – TomGrubb Sep 22 '17 at 16:51
  • $\begingroup$ Thank you:) Unfortunately the original question doesn't give any information about The domain and etc. $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 16:53
  • $\begingroup$ $f(x) = x^n \implies f(xy) = x^ny^n$ has this property. $\endgroup$ – Doug M Sep 22 '17 at 16:53
  • $\begingroup$ And $n=1/4$ is a possibility? $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 16:54
  • 5
    $\begingroup$ Without assuming e.g. continuity, there are also some very wild functions possible and we cannot obtain $f(2)$ $\endgroup$ – Hagen von Eitzen Sep 22 '17 at 16:55

Yes, the function is $f(x)=x^{\frac 14}$. You have shown that it works, but you need to show that it is the only function that works. I believe you should have been told that the function is continuous. As you say, you know that $f(9)=\sqrt 3, f(3)=\sqrt[4]3$. You can extend this to show that for any $x$ that is $3$ raised to a dyadic rational power (one of the form $\frac a{2^b}$) you have $f(x)=x^{\frac 14}$. Then argue that those numbers are dense in the positive reals so by continuity you have $f(x)=x^{\frac 14}$ for all positive reals.

  • $\begingroup$ Thank you:) Unfortunately the question doesn't give any information about continuity. But I guess we have to suppose it is because it is not a high level question. (İt is but) It was not well thought of, while being written. $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 16:59
  • $\begingroup$ We can define $g(x)=\log f(e^x)$ which makes the definining equation $g(x+y)=g(x)+g(y)$. This is known as Cauchy's functional equation with solution $g(x)=cx$ but without some further conditions there are crazy solutions. $\endgroup$ – Ross Millikan Sep 22 '17 at 17:53

Even assuming continuity and a domain of $\Bbb R$, there are at least two solutions: $$\tag1f(x)=\sqrt[4]{|x|} $$ and $$\tag2f(x)=\operatorname{sgn}(x)\cdot\sqrt[4]{|x|}. $$

What can we be sure of? From $f(0\cdot 81)=f(0)\cdot f(81)$ we conclude $f(0)=0$. From $f(1\cdot 81)=f(1)\cdot f(81)$ we conclude $f(1)=1$. On the other hand, if $x\ne 0$, then $f(x)\ne 0$ because $f(x)\cdot f(\frac{81}x)=f(81)\ne 0$.

From $f((-1)\cdot(-1))=f(-1)\cdot f(-1)$ we conclude $f(-1)=\pm1$ (and as the example solutions show, both are indeed possible). As $f(-x)=f(-1)f(x)$, we need only consider positive $x$, once we have picked a value for $f(-1)$. (note tha tin particular $|f(-x)|=|f(x)|$).

Now define $g(x)=\ln |f(e^x)|$. Then $$g(x+y)= \ln|f(e^{x+y})|=\ln|f(e^xe^y)|=\ln|f(e^x)f(e^y)|=\ln|f(x)|+\ln|f(y)|=g(x)+g(y).$$ This is a well-known equation, and under very mild additional conditions (such as: $g$ is continuous at at least one point, or: it is possible to write down an explicit expression for $g$), it follows that $g(x)=x\cdot g(1)$. Thus, for positive $x$, we find (under just as mild conditions) $$|f(x)|= e^{g(\ln x)}=e^{g(1)\ln x}=x^{g(1)}\qquad\text{for }x>0$$ and as $f(81)=3$, we must have $g(1)=\frac14$. The sign of $f$ might still flicker wildly, but invoking our "mild conditions" once again, we find that $(1)$ and $(2)$ are the only solutions.

  • $\begingroup$ Sorry for my ignorance $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 17:19
  • $\begingroup$ But what is sgn(x) $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 17:19
  • $\begingroup$ I think it's shorthand for the sign of x. aka -1 or +1 if you use the PARI/GP function. $\endgroup$ – user451844 Sep 22 '17 at 17:26
  • $\begingroup$ I see, İ thought it was something more complex,. Thank you:)) $\endgroup$ – Deniz Tuna Yalçın Sep 22 '17 at 18:24

If $f$ has domain over the integers, then let $g$ be any mapping of the primes into the reals. Let $x \geqslant 2$ be an integer. Then $f(x)=g(p_1)^{a_1}g(p_2)^{a_2}\cdots g(p_n)^{a_n},$ where $x={p_1}^{a_1}{p_2}^{a_2}\cdots {p_n}^{a_n}$ is the prime factorization of $x.$


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