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Question Number 216532 by sniper237 last updated on 11/Feb/25

$$Letf:{\mathbb{R}}_{+}\to \mathbb{R}suchasf\left(xy\right)=f\left(x\right)+f\left(y\right)$$$$1)Provethatfisderivableiff$$$$fisderivableatx=1.$$$$2)Provethatifso,f\left(x\right)={Log}_{a}x)$$$$whereaispositivevaluetoprecise$$

Answered by maths2 last updated on 11/Feb/25

$$f(x+y)=f\left(x\right)+f\left(y\right)..?$$$$$$

Commented by sniper237 last updated on 11/Feb/25

$$Sorryiforgot,{it}^{\prime }sf\left(xy\right)=f\left(x\right)+f\left(y\right)$$

Answered by maths2 last updated on 13/Feb/25

$$letDsetofderivablefunction$$$$iff\in D\Rightarrow fisderivableat1$$$$iffisderivableatx=1$$$$f\left(1\right)=0$$$$\frac{f\left(xy\right)-f\left(y\right)}{x-1}=\frac{f\left(x\right)-f\left(1\right)}{x-1}$$$$\underset{x\to 1}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=f^{\prime }\left(1\right)\in \mathbb{R}$$$$\underset{x\to 1}{\lim }\frac{f\left(xy\right)-f\left(y\right)}{x-1}=f^{\prime }\left(1\right),\mathrm{\forall }y\in \mathbb{R}$$$${\partial }_{x}f\left(xy\right){\mid }_{x=1}={yf}^{\prime }\left(y\right)\Rightarrow f^{\prime }\left(y\right)=\frac{f^{\prime }\left(1\right)}{y};\mathrm{\forall }y\in \mathbb{R}\Rightarrow f\left(y\right)\in D$$$$2)f^{\prime }\left(x\right)=\frac{1}{x}{f}^{\prime }\left(1\right);f^{\prime }\left(1\right)=a$$$$\Rightarrow f\left(x\right)=aln\left(x\right)+b;f\left(1\right)=0\Rightarrow b=0$$$$\Rightarrow f\left(x\right)=aln\left(x\right)$$$$leta=\frac{1}{ln\left(b\right)};b=e^{\frac{1}{f^{\prime }\left(1\right)}}$$$$\Rightarrow f\left(x\right)={ln}_b\left(x\right)$$

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