If $ \ln(x) f'(\ln(x))=3\ln^3x , \quad \forall x > 0$, prove $f(x)=x^3, x\in R$

Question

Let $f$ be a function, defined in $\mathbb{R}$ and differentiable in $\mathbb{R}$, with f(1)=1. If: $ \ln(x) f'(\ln(x))=3\ln^3x, \forall x > 0$, prove $f(x)=x^3, x\in R$. My try: $$f'(\ln(x))\frac{1}{x}= 3\ln^2(x)\frac{1}{x}, x\neq1 \Rightarrow (f(\ln(x)))'=(\ln^3x)' \Rightarrow f(\ln(x))=\ln^3x + c, \rightarrow c= 0$$ and if $$x=\ln(x) \Rightarrow f(x)=x^3$$ However, there are many spots where I think I need a better explanation. For example, I don't think I can just let $x=\ln(x)$ so freely.

Does anyone know what is needed here, or a better way? Thanks!

Answer 1

If you have$$(\forall x\in(0,\infty)):\log(x)f'(\log x)=3\log^3x,$$then, for each $x\in\Bbb R$,$$\log(e^x)f'(\log e^x)=3\log^3e^x;$$in other words, $xf'(x)=3x^3$, and therefore $f'(x)=3x^2$. So,$$f(x)=f(1)+\int_1^x3t^2\,\mathrm dt=1+x^3-1=x^3.$$

Answer 2

For $t\in \Bbb R$, let $x=e^t$. Then $$tf'(t)= \ln xf'(\ln x)=3\ln^3 x =3t^3.$$ Hence $$ f'(t)=3t^2\qquad\text{for }t\ne0.$$ By integration, it follows for $t>0$ that $$f(t)-f(1)=\int_1^t2u^2\,\mathrm du = t^3-1$$ and likewise for $t<0$ $$f(t)-f(-1)=\int_{-1}^t2u^2\,\mathrm du = t^3+1.$$ By continuity at $0$, we conclude that there exists a constant $C$ such that $$ f(t)=t^3+C$$ for all $t\in\Bbb R$. By the additional constraint $f(1)=1$, it follows that $C=0$.