The answer is $f(x) = x^\alpha$ only, where $\alpha \in \mathbb R_{>0}$. These are easily seen to work. Let $g(x) = \log f(e^x)$, so $g : \mathbb R_{>0} \to \mathbb R_{>0}$ and satisfies \[ 2x \le y \le 3x \implies 2g(x) \le g(y) \le 3g(x). \] We now claim $g$ is strictly increasing. Assume for contradiction $a > b$ but $g(a) \le g(b)$. Then for any integers $m$ and $n$ we have \[ \frac{1}{3^n} g(3^n a) \le g(a) \le g(b) \le \frac{1}{2^m} g(2^m b). \]Therefore, \[ g(2^m b) \ge 3 \cdot \frac{2^m}{3^{n+1}} g (3^n a). \]Now we select $m$ and $n$ so that $\frac{2^m}{3^{n+1}} > 1$ but $2^m b \le 3^{n+1} a$. This is possible since $\mathbb Q$ is dense: we just needed $m$ and $n$ such that \[ \log_2 3 < \frac{m}{n+1} < \log_2 \left( 3a/b \right). \]Thus $g$ is indeed strictly increasing. Now it remains to show $g(x) = g(1) x$ for every $x$. For any integers $m,n \ge 1$ satisfying $\frac{2^m}{3^n} x < 1$, we have \[ g(x) \le \frac{3^m}{2^n} g\left( \frac{2^m}{3^n} x \right) < \frac{3^m}{2^n} g(1). \]On the other hand, $\{ \frac{3^m}{2^n} \mid m,n \in \mathbb Z_{>0} \}$ is dense, since its base-2 logarithm is $m \log_2 3 - n$ and $\log_2 3 \notin \mathbb Q$. So $g(x) < (x + \varepsilon) g(1)$ for every $\varepsilon > 0$, hence $g(x) \le xg(1)$. The lower bound is exactly analogous.

My version, which never tries to mess with monotonicity or any other form of regularity until getting it all at once: Let $g$ be as in the previous post, and apply the log-log transformation again to get $h(u)=\log(g(\exp u))$ (Which log and exponential? It doesn't matter, as long as they're inverse to each other.) The condition then becomes that for any $u,v\in\mathbb{R}$ with $u+\log 2\le v\le u+\log 3$, $\log 2+h(u)\le h(v)\le \log 3 + h(u)$. Taking extreme values over $u$ for a particular $v$, this becomes $\log 2+\sup_{u\in [v-\log 3,v-\log 2]}h(u)\le h(v)\le \log 3 +\inf_{u\in [v-\log 3,v-\log 2]}h(u)$. Also, looking at the endpoints, $h(u)+\log 2\le h(u+\log 2)$ and $h(v)+\log 3\ge h(v+\log 3)$. Repeating this, $h(x)+j\log 2\le h(x+j\log 2)$ and $h(y)+k\log 3\ge h(y+k\log 3)$ for any real $x,y$ and positive integers $m,n$. Now, we want to show that $h(u)-u$ is constant. Suppose it takes two different values $h(x_0)-x_0=M>m=h(y_0)-y_0$. By the preceding, $h(x_0+j\log 2)-(x_0+j\log 2) \ge h(x_0)-x_0=M$ and $h(y_0+k\log 3)-(y_0+k\log 3)\le h(y_0)-y_0=m$. Since $\log 2$ and $\log 3$ are positive and incommensurable ($\frac{\log 3}{\log 2}$ is irrational), the set of possible values of $j\log 2-k\log 3$ for positive integers $j$ and $k$ is dense in $\mathbb{R}$ In particular, for any $\epsilon\in (0,\log 3-\log 2)$, there exist $j_{\epsilon}$ and $k_{\epsilon}$ with $\log 3-\log 2-\epsilon < (x_0+j_{\epsilon}\log 2)- (y_0+k_{\epsilon}\log 3) <\log 3-\log 2$. Let $X=x_0+j_{\epsilon}\log 2$ and $Y=y_0+k_{\epsilon}\log 3$ Then, for $V=\frac12(X+Y)+\frac12(\log 3+\log 2)$, $V-\log 3 \le Y<V-\log 3+\frac{\epsilon}{2}<V-\log 2-\frac{\epsilon}{2}<X<V-\log 2$. Now, \begin{align*}h(V) &\le \log 3 +\inf_{u\in [V-\log 3,V-\log 2]}h(u)\le \log 3 + h(Y)\\ &\le \log 3+Y+m\le \log 3+V-\log 3+\frac{\epsilon}{2} +m = V+m+\frac{\epsilon}{2}\end{align*}and similarly \begin{align*}h(V) &\ge \log 2+\sup_{u\in [V-\log 3,V-\log 2]}h(u) \ge \log 2 + h(X)\\ &\ge \log 2 + X + M\ge \log 2 + V-\log 2 -\frac{\epsilon}{2}+M=V+M-\frac{\epsilon}{2}\end{align*}Thus $M-\frac{\epsilon}{2}\le h(V)-V\le m+\frac{\epsilon}{2}$. For $\epsilon<M-m$, this is impossible. The only remaining possibility is for $h(u)-u$ to be constant on all of $\mathbb{R}$. Now, we have $h(u)=u+c$ for some real constant $c$ and all real $u$. Reversing the transformation, $g(t)=At$ for some positive constant $A$ and all positive $t$. Reversing the transformation again, $f(x)=x^A$ for some positive constant $A$ and all real $x$.

Here's a solution similar to v_Enhance's, but doesn't deal with monotonicity. As with the other solutions, let $g(x)=\ln f(e^x)$, so that its domain and range are $(0,\infty)$. We get that $2x\le y \le 3x \rightarrow 2g(x)\le g(y)\le 3g(x)$. Rewrite this as $2\le \frac{y}{x}\le 3\rightarrow 2\le \frac{g(y)}{g(x)}\le 3$. By writing it in this form, it is clear that we have $\frac{2^a}{3^b}\le \frac{y}{x}\le \frac{3^a}{2^b}\rightarrow \frac{2^a}{3^b}\le \frac{g(y)}{g(x)}\le \frac{3^a}{2^b}$ by multiplying sufficiently many inequalities of the form $2\le \frac{x_{i+1}}{x_i}\le 3$ where $x_0=x$ and $x_n=y$. From this, we get that $\frac{2^a}{3^b}\le y\le \frac{3^a}{2^b}\rightarrow \frac{2^a}{3^b}\le \frac{g(y)}{g(1)}\le \frac{3^a}{2^b}$. Suppose $y\neq \frac{g(y)}{g(1)}$. WLOG, $y<\frac{g(y)}{g(1)}$. Then we claim there are integers $a,b$ such that $y\le \frac{3^a}{2^b}\le \frac{g(y)}{g(1)}$, which proves the claim by contradiction since at least one of the inequalities must be strict. But this follows by taking the base two log: we want to find integers $a,b$ such that $a\log_2 3 - b$ is between $r$ and $s$ for any real $r,s$, which follows since $\log_2 3$ is irrational (we can take $a,b$ such that $|a\log_2 3 - b|<<|s-r|$ is very close to zero, and then scaling by an appropriate integer $k$ gets us the desired result).

We claim that the solutions are $f(x)=x^n$ for any $n>0$. It is easy to check that these work. Now suppose $f$ is a solution. Let \begin{align*}h:\mathbb{R}_{>0}&\to\mathbb{R}_{>0} \\ x&\mapsto \frac{1}{x}\log f(e^x). \end{align*}It is easy to see that this is well defined. Lemma 1: Suppose $a,b\in\mathbb{R}_{>0}$ such that $2a\le b\le 3a$. Then, $2h(a)\le \frac{b}{a}h(b)\le 3h(a)$. Proof of Lemma 1: Let $x=e^a$ and $y=e^b$. We see that $x^2\le y\le x^3$, so we have that \[f(x)^2\le f(y)\le f(x)^3.\]Taking the log, we see that \[2a h(a)\le bh(b)\le 3h(a),\]which is equivalent to what we wanted to show. $\blacksquare$ Note that lemma 1 quickly implies that $h(x)\ge h(x/2)$ (set $b=x$ and $a=x/2$) and $h(x)\ge h(3x)$ (set $b=3x$ and $a=x$). We now show that these are actually equalities. Lemma 2: $h(x)=h(x/2)$ and $h(x)=h(3x)$ for all $x\in\mathbb{R}_{>0}$. Proof of Lemma 2: Fix some particular value of $x$ that we are looking at (in particular when we say something is a constant in the following, it could change if $x$ changed). Let $m,n$ be positive integers that we can vary. If $h(x)-h(x/2)=C>0$, then \[h(x)=h(x/2)+C\ge C+h\left(\frac{3^n}{2^m}x\right),\]where the inequality follows from applying $\{h(y)\ge h(y/2)\}$ $m-1$ times and appying $\{h(y)\ge h(3y)\}$ $n$ times. Simarly, we see that if $h(x)-h(3x)=C>0$, then \[h(x)\ge C+h\left(\frac{3^n}{2^m}x\right).\]Either way, assuming statement of the lemma is false, then the above inequality holds for some positive constant $C$. Now, choose $n,m$ such that $3^n/2^m=2+\delta$ such that $0<\delta<1$. It is not hard to see that by varying $n,m$, we can make $\delta$ arbitrarily small. Lemma 1 with $b=(2+\delta)x$ and $a=x$ tells us that \[(2+\delta)h((2+\delta)x)\ge 2h(x).\]But since $h(x)\ge C+h((2+\delta)x)$, we have that \[(2+\delta)h((2+\delta)x)\ge 2C+2h((2+\delta)x),\]or \[2C\le \delta h((2+\delta)x).\]But again by lemma 1, we have that \[(2+\delta)h((2+\delta)x)\le 3h(x),\]so \[2C\le\frac{3\delta}{2+\delta}h(x).\]As we said, $\delta$ can be made arbitrarily small, the left side can be made arbitrarily small. But $C>0$, so we have the desired contradiction. $\blacksquare$ Note now that we have $h(x)=h(y)$ if $x/y\in S=\{3^n2^m:m,n\in\mathbb{Z}\}$. Choose any $a\in\mathbb{R}_{>0}$. Again, it is not hard to see that we can choose $\delta,\epsilon\in(0,1)$ such that $\delta+\epsilon$ is arbitrarily small and $\frac{2+\delta}{a},\frac{3-\epsilon}{a}\in S$. Lemma 1 implies that \[h(1)\le \frac{2+\delta}{2}h(2+\delta)\text{ and }h(1)\ge\frac{3-\epsilon}{3}h(3-\epsilon).\]Therefore, we have that \[\frac{3-\epsilon}{3}h(a)\le h(1)\le \frac{2+\delta}{2}h(a)\]for $\delta,\epsilon\in(0,1)$ such that $\delta+\epsilon$ is arbitrarily small. This clearly implies that $h(1)=h(a)$, so we have that $h(a)=h(1)=:n>0$ for all $a\in\mathbb{R}_{>0}$. Thus, \[\frac{1}{x}\log f(e^x)=n\implies f(e^x) = e^{nx}\]for all $x\in\mathbb{R}_{>0}$, so $f(x)=x^n$ for all $x\in\mathbb{R}_{>1}$, so $f$ is of the form that we claimed.

Solved with Vfire. Let $h(x)=\log f(e^x)$. The problem becomes: Find all $h:\mathbb{R}_{>0}\to \mathbb{R}_{>0}$ such that \[ 2\le \tfrac{b}{a} \le 3 \implies 2\le \tfrac{h(b)}{h(a)} \le 3\]for all $a,b>0$. We claim $h(x)=kx$ for some positive constant $k$. Claim: For any positive integers $m$ and $n$, we have \[ \frac{h(2^n/3^m)}{h(1)} \ge \frac{2^n}{3^m} \qquad \text{and} \qquad \frac{h(3^m/2^n)}{h(1)} \le \frac{3^m}{2^n}.\] Proof: We have Note $2\le \tfrac{2x}{x}\le 3$, so $h(2x)\ge 2h(x)$. Iterating, we get $h(2^nx)\ge 2^nh(x)$ for all positive integers $n$. Also, $2\le \tfrac{y}{y/3}\le 3$, so $\tfrac{h(y)}{h(y/3)}\le 3$, so $h(\tfrac{1}{3}y)\ge \tfrac13 h(y)$. Iterating this process, we get $h(\tfrac{1}{3^m}y)\ge \tfrac{1}{3^m}h(y)$ for all positive integers $m$. Plugging in $x=y/3^m$ into the first, we get \[ h\left(\frac{2^n}{3^m}y\right) \ge 2^nh\left(\frac{y}{3^m}\right) \ge \frac{2^n}{3^m}h(y), \]so plugging in $y=1$ proves the first part of the claim. The second part is easy to show similarly by reciprocating everything. $\blacksquare$ Claim: $h$ is strictly increasing. Proof: Suppose $x>y$ and $h(x)\le h(y)$. Then for any positive integers $d$ and $e$, we have $h(2^dy)\ge 2^dy$ and $f(3^ex)\le 3^eh(x)$, so \[ \frac{h(3^ex)}{3^e} \le h(x) \le h(y) \le \frac{h(2^dy)}{2^d} \implies \frac{h(2^dy)}{h(3^ex)} \ge \frac{2^d}{3^e}. \]If we can choose $d$ and $e$ such that $\tfrac{2^dy}{3^ex}\le 3$ but $\tfrac{2^d}{3^e}>3$, then we have a contradiction. Since $\tfrac{y}{x}<1$ and $\{2^d/3^e \mid d,e\in \mathbb{N}\}$ is dense in $\mathbb{Q}$, we can choose suitable $d$ and $e$. $\blacksquare$ Fix a real $x$. Since $\{2^d/3^e \mid d,e\in \mathbb{N}\}$ is dense in $\mathbb{Q}$, we can approximate $x$ by $2^n/3^m$ for some $m$ and $n$ and by $3^{m'}/2^{n'}$ for some $m'$ and $n'$. Since $h$ is increasing, bounding $x$ between $2^n/3^m$ and $3^{m'}/2^{n'}$ bounds $h(x)$ between $h(2^n/3^m)/h(1)$ and $h(3^{m'}/2^{n'})/h(1)$. Making these arbitrarily close gives $h(x)/h(1)=x$, as desired.

Nice problem for oral examination Polytechnique year $\geq 2021$

Quote: We need $h$ increasing to make sure $h$ isn't wacky so that the interval bounding argument works. It's hard to get all the inequality signs to work, but I think the above works. I don't think $h$ increasing is necessary? (Though this is probably the same thing anyway.) Assume $h(1) = 1$ in what follows (by scaling). After you get \[ h\left(\frac{2^n}{3^m}\right)\geq \frac{2^n}{3^m}\quad\text{and}\quad h\left(\frac{3^m}{2^n}\right)\leq \frac{3^m}{2^n}, \]I think you can finish in the following way. Let $r\in(1,\infty)$ be arbitrary. Since $\{2^n/3^m\}_{(m,n)\in\mathbb N^2}$ is dense in $\mathbb Q$, there exists an increasing sequence in $S$ converging to $\tfrac r2$; that is, $\tfrac{2^n}{3^m}\nearrow \tfrac r2$. Thus \[ h(r) \geq 2h\left(\frac{r}{2}\right)\geq 2\cdot\frac{2^n}{3^m}\nearrow r. \]This means $h(r) \geq r$. Using the upper bound instead yields $h(r) \leq r$, so $h(r) = r$ and we may conclude. $\blacksquare$

The answer is $f(x)\equiv x^c$ for some $c\in \mathbb{R}_{>0}$. All logs are done in the same base (for example, $e$) Let $g(\log x)=\log(f(x))$. Then $g$ is from $R^+ \to R^+$. Let $a=\log x, b=\log y$. If $2a\le b\le 3a$ then $2g(a)<g(b)<3g(a)$ Let $h(\log x)=\log g(x)$. Then $h$ is from $R \to R$. Let $c=\log a, d=\log b, \alpha=\log 2, \beta=\log 3$, then if $d-c\in [\alpha,\beta]$ then $h(d)-h(c)\in [\alpha,\beta]$. This implies for positive integer $t,u$, $h(x+t\alpha)-h(x)\ge t\alpha$ and $h(x+u\beta)-h(x) \le u\beta$ If $t,u$ are large, $h(x+t\alpha+\epsilon)-h(x)\ge t\alpha, h(x+u\beta-\epsilon)-h(x)\le u\beta$ still holds. I claim $h(x)-h(y)=x-y$ by proving $h(x)-h(y)\ge x-y$, and $h(x)-h(y)\le x-y$ is similar. This proves $h(x)-x$ is constant, so $h(x)=x+C$, $g(x)=e^{h(\log x)}=e^{\log x+C}=cx$ where $c=e^C$, and $f(x)=e^{g(\log x)}=e^{c\log x} = x^c$ Select $t,u$ large such that $h(x)-h(y)<t\alpha-u\beta<x-y$, which exist because $\frac{\beta}{\alpha} \notin \mathbb{Q}$. Select $\epsilon_1, \epsilon_2>0$ such that $x-t\alpha-\epsilon_1 = y-u\beta + \epsilon_2=z$. Then $h(z)=h(x-t\alpha-\epsilon_1)\le h(x)-t\alpha < h(y)-u\beta \le h(y-u\beta+\epsilon)=h(z)$

The answer is $f(x)=x^c$ for some real $c>0$, which clearly works. We have the following claim. Claim: Let $r>0$ be a real number. As $i,j$ range across the nonnegative integers, $\frac{3^j}{2^i}r-1$ takes on arbitrarily small positive values. Proof: By exponentiation, it suffices to show that $\frac{3^j}{2^i}$ takes on values arbitrarily close but still larger than $1$. By taking a logarithm, it suffices to show that $j\ln(3)-i\ln(2)$ takes on arbitrarily small positive values. This follows because it is dense in $\mathbb{R}$. $\blacksquare$ Now suppose that $\log_x(f(x))$ is nonconstant, and pick $a,b$ such that $f(a)=a^p$ and $f(b)=a^q$ with $p>q$. By repeatedly applying the problem statement we have $f\left(a^{2^i}\right) \geq a^{2^ip}$ and $f\left(b^{3^j}\right) \leq b^{3^jq}$. Then pick some $i,j$ such that $\frac{3^j\ln(b)}{2^i\ln(a)}$ is close but still greater than $1$. Since $a^{2^i} \leq b^{3^j} \leq a^{3\cdot 2^{i-1}}$, it follows that $b^{3^jq}\geq a^{2^ip}$, hence $\frac{3^j\ln(b)}{2^i\ln(a)} \geq \frac{p}{q}$, which is a contradiction. Thus $\log_x(f(x))$ is nonconstant, and clearly only positive reals work for this constant value, finishing the problem. $\blacksquare$

Exponent spam. Claim: For any $a$ and $b$ and $\varepsilon > 0$, there exist infinitely many positive integers $m$ and $n$ such that \[ b^{3^m \cdot (3 - \varepsilon)} \le a^{2^n} \le b^{3^m \cdot 3} \]Proof. Take $\log$ two times then follows as $\frac{\log 3}{\log 2}$ is irrational. $\blacksquare$ Claim: There can't exist $a, b$ such that $f(a) > a$ and $f(b) \le b$. Proof. Note that it follows from this that $f(b^{3^m}) \le b^{3^m}$. Similarily, if we let $c = \frac{f(a)}{a}$, then it follows that $f(a^{2^n}) \ge a^{2^n} \cdot c^{2^n}$ by the claim. Take $\varepsilon$ such that $c^{2^n} > b^{\log_b(bc^{2^n}) \cdot \varepsilon}$ for large $n$ then gives that for some $m, n$ \[ c^{2^m} \cdot a^{2^m} \le b^{3^m \cdot 3} \]a contradiction. $\blacksquare$ Now, note that $f$ satisfies this FE, then $f^k$ does as well. The above claim implies that for any $a, b$, we have that $\frac{f(a)^{k}}{a} = 1 \iff \frac{f(b)^k}{b} = 1$, from which we get $f(x) = x^k$ for $k > 0$.