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Let, $f$: $\mathbb{R^+}$$\rightarrow$$\mathbb{R}$ be a continuous function satisfying $f(xy)=f(x)+f(y)$. Prove that, $f(x)=c\log x$ for some $c>0$.

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  • $\begingroup$ See the way which was described here. $\endgroup$ – mrs Dec 22 '13 at 6:18
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    $\begingroup$ Define $g(x)=f(e^x)$ and check that $g(x)=cx$ for some $x$. $\endgroup$ – Hanul Jeon Dec 22 '13 at 6:36
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It is obvious that the (continuous) function $\theta:\mathbb{R}^+\to \mathbb{R},\ \theta(x)=0$ satisfies the functional equation $$\tag{*} f(xy)=f(x)+f(y). $$ Let $f \not\equiv \theta$ be a solution of (*). Then we have $$ f(1)=f(1)+f(1), $$ i.e. $$\tag{1} f(1)=0 $$ By induction we have $$\tag{2} f(x^n)=nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{N}. $$ Thanks to (1), we have $$\tag{3} f(x^{-1})=f(x^{-1})+f(x)-f(x)=f(x^{-1}x)-f(x)=f(1)-f(x)=-f(x)\quad \forall x \in \mathbb{R}^+. $$ Combining (2) and (3), we deduce that $$\tag{4} f(x^n)=nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{Z}. $$ Thanks to (4), we have $$ f(x^{1/n})=\frac1nf((x^{1/n})^n)=\frac1nf(x) \quad \forall x \in \mathbb{R}^+,\, n \in \mathbb{N}. $$ Thus $$ f(x^{m/n})=\frac{m}{n}f(x)\quad \forall x \in \mathbb{R}^+,\,m\in \mathbb{Z},\, n\in \mathbb{N}, $$ in other words we have $$ f(x^q)=qf(x)\quad \forall x \in \mathbb{R}^+,\,q\in \mathbb{Q}. $$ Since $f$ is continuous, and $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $$ f(x^\alpha)=\alpha f(x)\quad \forall x \in \mathbb{R}^+,\,\alpha\in \mathbb{R}. $$ Since $f\not\equiv\theta$, there is some $z \in \mathbb{R}^+\setminus\{1\}$ such that $f(z)\ne 0$. W.l.o.g. assume $1<z<e$ and $f(z)>0$ (otherwise we replace $z$ by $z^{-1}$). Let $s=\frac{1}{\ln z} \in \mathbb{R}^+$. Then, $z^s=e$ and for every $x \in \mathbb{R}^+$ we have $$ f(x)=f(e^{\ln x})=f(e)\ln x=f(z^s)\ln x=sf(z)\ln x=c\ln x, $$ where $c=s\ln z$.

Remark: It is obvious that the constant $c$ need not be positive.

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First show that $f\left(x^{\frac{m}{n}}\right) = \frac{m}{n}f(x)$ for any rational number $\frac{m}{n}$. Then use continuity to get $f(x^\alpha) = \alpha f(x)$ for all $\alpha \in \mathbb R$.

Now note that $f(1) = 0$. Suppose $f(k) = 0$ for some $k > 1$. Then for all $x$ we have $f(x) = f(k^\alpha) = \alpha f(k) = 0$ for some $\alpha$. Hence $f(x) = c\log(x)$ with constant $c = 0$.

Otherwise suppose, without loss of generality, that $f(k) > 0$ for some $k > 1$ (if $f(k) < 0$, then consider the function $-f$ for the following). Then there is some $\alpha > 0$ so that $f(k^\alpha) = \alpha f(k) = 1$. Let $b = k^\alpha$.

Then for any $x$ we have $x = b^\alpha$ where $\alpha = \frac{\log(x)}{\log(b)}$. So $f(x) = f(b^\alpha) = \alpha f(b) = \alpha = \frac{\log(x)}{\log(b)}$. Hence $f(x) = c\log(x)$ with constant $c = \frac{1}{\log(b)}$.

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