Question-159794

Question Number 159794 by 0731619 last updated on 21/Nov/21

Commented by cortano last updated on 21/Nov/21

(d/dx) (∫ 2^x dx ) = (d/dx)(x^2 ) ⇒2^x = 2x ⇒ 2^(x−1) = x ⇒(x−1)ln (2)= ln (x) ⇒x ln (2)−ln (x)= ln (2)

$$\:\frac{{d}}{{dx}}\:\left(\int\:\mathrm{2}^{{x}} \:{dx}\:\right)\:=\:\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{2}^{{x}} \:=\:\mathrm{2}{x} \\ $$$$\Rightarrow\:\mathrm{2}^{{x}−\mathrm{1}} \:=\:{x} \\ $$$$\Rightarrow\left({x}−\mathrm{1}\right)\mathrm{ln}\:\left(\mathrm{2}\right)=\:\mathrm{ln}\:\left({x}\right) \\ $$$$\Rightarrow{x}\:\mathrm{ln}\:\left(\mathrm{2}\right)−\mathrm{ln}\:\left({x}\right)=\:\mathrm{ln}\:\left(\mathrm{2}\right) \\ $$$$ \\ $$

Commented by mr W last updated on 21/Nov/21

i don′t think it′s correct sir. to solve an equation f(x)=g(x), it is to find the value of x, say a, which fulfills the equation f(a)=g(a). here a is a concrete value, and you can not treat it as a variable and derivate w.r.t. to a. example: to solve x^3 +x^2 +1=3x you can not do like this: (d/dx)(x^3 +x^2 +1)=(d/dx)(3x) 3x^2 +2x=3 3x^2 +2x−3=0 x=((−1±(√(10)))/3)

$${i}\:{don}'{t}\:{think}\:{it}'{s}\:{correct}\:{sir}. \\ $$$${to}\:{solve}\:{an}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right), \\ $$$${it}\:{is}\:{to}\:{find}\:{the}\:{value}\:{of}\:{x},\:{say}\:{a}, \\ $$$${which}\:{fulfills}\:{the}\:{equation} \\ $$$${f}\left({a}\right)={g}\left({a}\right). \\ $$$${here}\:{a}\:{is}\:{a}\:{concrete}\:{value},\:{and}\:{you}\: \\ $$$${can}\:{not}\:{treat}\:{it}\:{as}\:{a}\:{variable}\: \\ $$$${and}\:{derivate}\:{w}.{r}.{t}.\:{to}\:{a}. \\ $$$$ \\ $$$${example}:\:{to}\:{solve}\:{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}=\mathrm{3}{x} \\ $$$${you}\:{can}\:{not}\:{do}\:{like}\:{this}: \\ $$$$\frac{{d}}{{dx}}\left({x}^{\mathrm{3}} +{x}^{\mathrm{2}} +\mathrm{1}\right)=\frac{{d}}{{dx}}\left(\mathrm{3}{x}\right) \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}=\mathrm{3} \\ $$$$\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}=\mathrm{0} \\ $$$${x}=\frac{−\mathrm{1}\pm\sqrt{\mathrm{10}}}{\mathrm{3}} \\ $$

Commented by amin96 last updated on 21/Nov/21

$$ \\ $$What program do you use to draw graphics?

Commented by mr W last updated on 21/Nov/21

$${whom}\:{do}\:{you}\:{ask}?\:{which}\:{graphics}\:{do} \\ $$$${you}\:{mean}? \\ $$

Commented by amin96 last updated on 21/Nov/21

$$ \\ $$I mean the funksion graphics. Which app are you watching?

Commented by Tony6400 last updated on 21/Nov/21

$$ \\ $$

Commented by Tony6400 last updated on 21/Nov/21

$$ \\ $$

Commented by Tony6400 last updated on 21/Nov/21

Take u=2^x so lnu=xln2 (u^′ /u)=ln2 and u^′ =uln2>>>(du/dx)=2^x ln2>>u=ln2∫2^x dx so (u/(ln2))=∫2^x dx=(2^x /(ln2)) so (2^x /(ln2))=x^2 >>>2^x =x^2 ln2 No solution

$${Take}\:{u}=\mathrm{2}^{{x}} \:{so}\:{lnu}={xln}\mathrm{2} \\ $$$$\frac{{u}^{'} }{{u}}={ln}\mathrm{2}\:{and}\:{u}^{'} ={uln}\mathrm{2}>>>\frac{{du}}{{dx}}=\mathrm{2}^{{x}} {ln}\mathrm{2}>>{u}={ln}\mathrm{2}\int\mathrm{2}^{{x}} {dx} \\ $$$${so}\:\frac{{u}}{{ln}\mathrm{2}}=\int\mathrm{2}^{{x}} {dx}=\frac{\mathrm{2}^{{x}} }{{ln}\mathrm{2}} \\ $$$${so}\:\frac{\mathrm{2}^{{x}} }{{ln}\mathrm{2}}={x}^{\mathrm{2}} >>>\mathrm{2}^{{x}} ={x}^{\mathrm{2}} {ln}\mathrm{2} \\ $$$${No}\:{solution} \\ $$

Answered by mr W last updated on 21/Nov/21

f(x)=∫2^x dx=∫e^(xln 2) dx=(1/(ln 2))∫e^(xln x) d(xln 2) =(1/(ln 2))e^(xln 2) +C=(2^x /(ln 2))+C g(x)=x^2 so the equation f(x)=g(x) is (2^x /(ln 2))+C=x^2 it has no unique solution due to unknown constant C.

$${f}\left({x}\right)=\int\mathrm{2}^{{x}} {dx}=\int{e}^{{x}\mathrm{ln}\:\mathrm{2}} {dx}=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}\int{e}^{{x}\mathrm{ln}\:{x}} {d}\left({x}\mathrm{ln}\:\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{ln}\:\mathrm{2}}{e}^{{x}\mathrm{ln}\:\mathrm{2}} +{C}=\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}+{C} \\ $$$${g}\left({x}\right)={x}^{\mathrm{2}} \\ $$$${so}\:{the}\:{equation}\:{f}\left({x}\right)={g}\left({x}\right)\:{is} \\ $$$$\frac{\mathrm{2}^{{x}} }{\mathrm{ln}\:\mathrm{2}}+{C}={x}^{\mathrm{2}} \\ $$$${it}\:{has}\:{no}\:{unique}\:{solution}\:{due}\:{to}\: \\ $$$${unknown}\:{constant}\:{C}. \\ $$

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