Menu Close

A-dx-x-n-A-ln-x-n-ln-n-ln-n-ln-n-n-ln-n-n-n-A-dx-x-n-ln-n-1-For-a-b-c-i-if-a-gt-b-c-gt-1-ii

Tinku Tara 0 Comments
Pin




Question Number 3017 by Filup last updated on 03/Dec/15
A=∫_μ ^( μ+ε) (dx/(x+n)) A=ln(x+n) ∣_μ ^(μ+ε) =ln(μ+ε+n)−ln(μ+n) =ln(((μ+ε+n)/(μ+n))) =ln(((μ+n)/(μ+n))+(ε/(μ+n))) ∴A=∫_μ ^( μ+ε) (dx/(x+n))=ln((ε/(μ+n))+1) For (a/b)=c i) if a>b, c>1 ii) if a<b, c<1 iii) if a=b, c=1 Case (i) ε>μ+n (ε/(μ+n))>1 ∴A>ln(2) Case (ii) ε<μ+n (ε/(μ+n))<1 ∴A<ln(2) Case (iii) ε=μ+n (ε/(μ+n))=1 A=ln(2) What other observations can you find regarding similar logarithmic functions?
$${A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}} \\ $$$$ \\ $$$${A}=\mathrm{ln}\left({x}+{n}\right)\:\mid_{\mu} ^{\mu+\epsilon} \\ $$$$=\mathrm{ln}\left(\mu+\epsilon+{n}\right)−\mathrm{ln}\left(\mu+{n}\right) \\ $$$$=\mathrm{ln}\left(\frac{\mu+\epsilon+{n}}{\mu+{n}}\right) \\ $$$$=\mathrm{ln}\left(\frac{\mu+{n}}{\mu+{n}}+\frac{\epsilon}{\mu+{n}}\right) \\ $$$$ \\ $$$$\therefore{A}=\int_{\mu} ^{\:\mu+\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\epsilon}{\mu+{n}}+\mathrm{1}\right) \\ $$$$ \\ $$$$\mathrm{For}\:\:\frac{{a}}{{b}}={c} \\ $$$$\left.{i}\right)\:\mathrm{if}\:\:{a}>{b},\:{c}>\mathrm{1} \\ $$$$\left.{ii}\right)\:\mathrm{if}\:\:\:{a}<{b},\:{c}<\mathrm{1} \\ $$$$\left.{iii}\right)\:\mathrm{if}\:\:{a}={b},\:{c}=\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Case}\:\left({i}\right) \\ $$$$\epsilon>\mu+{n} \\ $$$$\frac{\epsilon}{\mu+{n}}>\mathrm{1} \\ $$$$\therefore{A}>\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{Case}\:\left({ii}\right) \\ $$$$\epsilon<\mu+{n} \\ $$$$\frac{\epsilon}{\mu+{n}}<\mathrm{1} \\ $$$$\therefore{A}<\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{Case}\:\left({iii}\right) \\ $$$$\epsilon=\mu+{n} \\ $$$$\frac{\epsilon}{\mu+{n}}=\mathrm{1} \\ $$$${A}=\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$ \\ $$$$\mathrm{What}\:\mathrm{other}\:\mathrm{observations}\:\mathrm{can}\:\mathrm{you}\:\mathrm{find} \\ $$$$\mathrm{regarding}\:\mathrm{similar}\:\mathrm{logarithmic}\:\mathrm{functions}? \\ $$
Commented by Filup last updated on 03/Dec/15
Maybe look at cases of ∫_μ ^( με) , ∫_μ ^( μ^ε ) , etc.
$$\mathrm{Maybe}\:\mathrm{look}\:\mathrm{at}\:\mathrm{cases}\:\mathrm{of}\:\int_{\mu} ^{\:\mu\epsilon} ,\:\int_{\mu} ^{\:\mu^{\epsilon} } ,\:{etc}. \\ $$
Commented by 123456 last updated on 03/Dec/15
μ>−n∨μ+ε<−n
$$\mu>−{n}\vee\mu+\epsilon<−{n} \\ $$
Commented by Filup last updated on 03/Dec/15
Please elaborate?
$$\mathrm{Please}\:\mathrm{elaborate}? \\ $$
Answered by Filup last updated on 03/Dec/15
∫_μ ^(με) (dx/(x+n))=ln(((με+n)/(μ+n))) =ln(((ε+(n/μ))/(1+(n/μ)))) k=(n/μ) A=ln(((ε+k)/(1+k))) Similar to original post: ε>1, A>0 ε<1, A<0 ε=1, A=0
$$\int_{\mu} ^{\mu\epsilon} \:\frac{{dx}}{{x}+{n}}=\mathrm{ln}\left(\frac{\mu\epsilon+{n}}{\mu+{n}}\right) \\ $$$$=\mathrm{ln}\left(\frac{\epsilon+\frac{{n}}{\mu}}{\mathrm{1}+\frac{{n}}{\mu}}\right) \\ $$$$ \\ $$$${k}=\frac{{n}}{\mu} \\ $$$${A}=\mathrm{ln}\left(\frac{\epsilon+{k}}{\mathrm{1}+{k}}\right) \\ $$$$ \\ $$$$\mathrm{Similar}\:\mathrm{to}\:\mathrm{original}\:\mathrm{post}: \\ $$$$\epsilon>\mathrm{1},\:{A}>\mathrm{0} \\ $$$$\epsilon<\mathrm{1},\:{A}<\mathrm{0} \\ $$$$\epsilon=\mathrm{1},\:{A}=\mathrm{0} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *