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

Let f :R_+ →R such as f(xy)=f(x)+f(y) 1) Prove that f is derivable iff f is derivable at x=1. 2) Prove that if so, f(x)=Log_a x) where a is positive value to precise

$${Let}\:{f}\::\mathbb{R}_{+} \rightarrow\mathbb{R}\:{such}\:{as}\:{f}\left({xy}\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$$$\left.\mathrm{1}\right)\:{Prove}\:{that}\:\:{f}\:{is}\:{derivable}\:{iff}\:\: \\ $$$${f}\:{is}\:{derivable}\:{at}\:{x}=\mathrm{1}. \\ $$$$\left.\mathrm{2}\left.\right)\:{Prove}\:{that}\:{if}\:{so},\:{f}\left({x}\right)={Log}_{{a}} {x}\right)\: \\ $$$${where}\:{a}\:{is}\:{positive}\:{value}\:{to}\:{precise} \\ $$

Answered by maths2 last updated on 11/Feb/25

f(x+y)=f(x)+f(y)..?

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

Commented by sniper237 last updated on 11/Feb/25

Sorry i forgot , it′s f(xy)=f(x)+f(y)

$${Sorry}\:{i}\:{forgot}\:,\:{it}'{s}\:{f}\left({xy}\right)={f}\left({x}\right)+{f}\left({y}\right) \\ $$

Answered by maths2 last updated on 13/Feb/25

let D set of derivable function if f∈D⇒f is derivable at 1 if f is derivable at x=1 f(1)=0 ((f(xy)−f(y))/(x−1))=((f(x)−f(1))/(x−1)) lim_(x→1) ((f(x)−f(1))/(x−1))=f′(1)∈R lim_(x→1) ((f(xy)−f(y))/(x−1))=f′(1),∀y∈R ∂_x f(xy)∣_(x=1) =yf′(y)⇒f′(y)=((f′(1))/y);∀y∈R⇒f(y)∈D 2) f′(x)=(1/x)f′(1);f′(1)=a ⇒f(x)=aln(x)+b;f(1)=0⇒b=0 ⇒f(x)=aln(x) let a=(1/(ln(b)));b=e^(1/(f′(1))) ⇒f(x)=ln_b (x)

$${let}\:{D}\:{set}\:{of}\:{derivable}\:{function} \\ $$$${if}\:{f}\in{D}\Rightarrow{f}\:{is}\:{derivable}\:{at}\:\mathrm{1} \\ $$$${if}\:{f}\:{is}\:{derivable}\:{at}\:{x}=\mathrm{1} \\ $$$${f}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\frac{{f}\left({xy}\right)−{f}\left({y}\right)}{{x}−\mathrm{1}}=\frac{{f}\left({x}\right)−{f}\left(\mathrm{1}\right)}{{x}−\mathrm{1}} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{f}\left({x}\right)−{f}\left(\mathrm{1}\right)}{{x}−\mathrm{1}}={f}'\left(\mathrm{1}\right)\in\mathbb{R} \\ $$$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{f}\left({xy}\right)−{f}\left({y}\right)}{{x}−\mathrm{1}}={f}'\left(\mathrm{1}\right),\forall{y}\in\mathbb{R} \\ $$$$\partial_{{x}} {f}\left({xy}\right)\mid_{{x}=\mathrm{1}} ={yf}'\left({y}\right)\Rightarrow{f}'\left({y}\right)=\frac{{f}'\left(\mathrm{1}\right)}{{y}};\forall{y}\in\mathbb{R}\Rightarrow{f}\left({y}\right)\in{D} \\ $$$$\left.\mathrm{2}\right)\:{f}'\left({x}\right)=\frac{\mathrm{1}}{{x}}{f}'\left(\mathrm{1}\right);{f}'\left(\mathrm{1}\right)={a} \\ $$$$\Rightarrow{f}\left({x}\right)={aln}\left({x}\right)+{b};{f}\left(\mathrm{1}\right)=\mathrm{0}\Rightarrow{b}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)={aln}\left({x}\right) \\ $$$${let}\:{a}=\frac{\mathrm{1}}{{ln}\left({b}\right)};{b}={e}^{\frac{\mathrm{1}}{{f}'\left(\mathrm{1}\right)}} \\ $$$$\Rightarrow{f}\left({x}\right)={ln}_{{b}} \left({x}\right) \\ $$

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