Let $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ be a function satisfying $$f(x)f\big(yf(x)\big)=f(x+y),\forall x,y\in\mathbb{R}^+$$ If $f(1)=\frac{1}{152}$, evaluate $f(4)$.
By inspection, we can see $f(x)=\frac{1}{151x+1}$ is a solution, from which we can easily get the answer. But how can we show that this is the only solution?
Here is my work:
Because $f$ is the reciprocal of a linear function, it would probably help to define $g(x)=\frac{1}{f(x)}$ (note that this is well defined as we are working in $\mathbb{R}^+$). Then the given equation becomes $$g(x)g\left(\frac{y}{g(x)}\right)=g(x+y)$$
If we take $y=g(x)$, then this becomes $$g(x)g(1)=g\big(x+g(x)\big)\implies 152g(x)=g\big(x+g(x)\big)$$
Not sure where to go from here. Any thoughts?
I'm going to proceed with your $g(x) = 1/f(x)$, starting with this $$g(x)g\left(\frac{y}{g(x)}\right)=g(x+y)$$ but then re-setting in it $y = g(x)y_1$ (possible for any positive $y_1$) to get this basic equation $$\tag1 g(x)g(y)=g(x+g(x)y)$$ (in which I renamed $y_1$ back to $y$). Applying this first with $x=1$ and then with some other $x = w > 1$, letting $B := g(w)$, gives $$ g(1+152y) = 152g(y)\\ g(w+By) = Bg(y) $$ Suppose $B < 152$. We can then pick a positive $y$ such that $1+152y = w + By$ so that the left-hand sides are equal, but the right-hand sides are not (since $g(y) > 0$) - contradiction. So $w > 1$ implies $g(w) \geq 152$.
Now consider $u = w + B(1+152y_0)$ and $v = 1+152(w+By_0)$ for some arbitrary $y_0 > 0$: $$ g(u) = g(w + B + 152By_0) = Bg(1+152y_0) = 152Bg(y_0)\\ g(v) = g(1+ 152w +152By_0) = 152g(w+By_0) = 152Bg(y_0) $$ Since the right-hand-sides are equal, $g(u) = g(v)$. This means that either $w+B = 1+152w$, meaning that $$\tag2 g(w) = B = 1+151w $$ as we wanted to prove, or, $g(u) = g(v)$ with $u \neq v$. We derive a contradiction in the latter case. WLOG, suppose $u < v$. Then setting in (1) $x = u, y = (v-u)/g(u) =: t$, we get $$ g(u)g(t) = g(u + g(u)t) = g(v)\\ \therefore g(t) = 1 $$ Now set in (1) $x = t$, to get $g(t+y) = g(y)$ and since $g(t) = 1$, it follows by induction that $g(nt) = 1$ for any positive integer $n$. By choosing a sufficiently large $n$, we can make $nt > 1, g(nt) = 1$ which is a contradiction with the previous result that $w > 1 \Rightarrow g(w) \geq 152$. This leaves (2) as the unique solution.