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Question Number 25972 by mondodotto@gmail.com last updated on 17/Dec/17

Answered by rita1608 last updated on 17/Dec/17

differentiating on both sides we get (d/dx)(logx^2 )=(1/(25)) (1/x^2 )×2x=(1/(25)) (2/x)=(1/(25)) x=50

$${differentiating}\:{on}\:{both}\:{sides}\: \\ $$$${we}\:{get}\:\frac{{d}}{{dx}}\left({logx}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} }×\mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$$\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{{x}}=\frac{\mathrm{1}}{\mathrm{25}} \\ $$$${x}=\mathrm{50} \\ $$

Commented by prakash jain last updated on 17/Dec/17

rita1608, f(x)=g(x) ⇒given equation you are solving an equation f ′(x)=g′(x) if f′(x)=g′(x) at x=x_0 it does not imply f(x)=g(x) at x=x_0 .

$$\mathrm{rita1608}, \\ $$$${f}\left({x}\right)={g}\left({x}\right)\:\Rightarrow{given}\:{equation} \\ $$$$\mathrm{you}\:\mathrm{are}\:\mathrm{solving}\:\mathrm{an}\:\mathrm{equation}\: \\ $$$${f}\:'\left({x}\right)={g}'\left({x}\right) \\ $$$$ \\ $$$$\mathrm{if}\:{f}'\left({x}\right)={g}'\left({x}\right)\:\mathrm{at}\:{x}={x}_{\mathrm{0}} \:\mathrm{it}\:\mathrm{does} \\ $$$$\mathrm{not}\:\mathrm{imply}\:{f}\left({x}\right)={g}\left({x}\right)\:\mathrm{at}\:{x}={x}_{\mathrm{0}} . \\ $$

Answered by mrW1 last updated on 17/Dec/17

if x>0: ⇒2log x=(x/(25)) ⇒((ln x)/(ln 10))=(x/(50)) ⇒ln x=((ln 10)/(50))×x=ax with a=((ln 10)/(50)) ⇒x=e^(ax) ⇒xe^(−ax) =1 ⇒(−ax)e^(−ax) =−a ⇒−ax=W(−a) with W=Lambert W−function ⇒x=−((W(−a))/a)=−((W(−((ln 10)/(50))))/((ln 10)/(50))) if x<0: ⇒2log (−x)=(x/(25)) ⇒((ln (−x))/(ln 10))=(x/(50)) ⇒ln (−x)=((ln 10)/(50))×x=ax with a=((ln 10)/(50)) ⇒−x=e^(ax) ⇒(−ax)e^(−ax) =a ⇒−ax=W(a) ⇒x=−((W(a))/a)=−((W(((ln 10)/(50))))/((ln 10)/(50))) all solutions: ⇒x=−((W(±((ln 10)/(50))))/((ln 10)/(50)))= { ((−((W(0.046052))/(0.046052))=−((0.044066)/(0.046052))=−0.9569)),((−((W(−0.046052))/(0.046052))= { ((−((−0.048332)/(0.046052))=1.0495)),((−((−4.6052)/(0.046052))=100)) :})) :}

$${if}\:{x}>\mathrm{0}: \\ $$$$\Rightarrow\mathrm{2log}\:{x}=\frac{{x}}{\mathrm{25}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:{x}}{\mathrm{ln}\:\mathrm{10}}=\frac{{x}}{\mathrm{50}} \\ $$$$\Rightarrow\mathrm{ln}\:{x}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}×{x}={ax}\:{with}\:{a}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}} \\ $$$$\Rightarrow{x}={e}^{{ax}} \\ $$$$\Rightarrow{xe}^{−{ax}} =\mathrm{1} \\ $$$$\Rightarrow\left(−{ax}\right){e}^{−{ax}} =−{a} \\ $$$$\Rightarrow−{ax}=\mathbb{W}\left(−{a}\right)\:{with}\:{W}={Lambert}\:{W}−{function} \\ $$$$\Rightarrow{x}=−\frac{{W}\left(−{a}\right)}{{a}}=−\frac{{W}\left(−\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}} \\ $$$${if}\:{x}<\mathrm{0}: \\ $$$$\Rightarrow\mathrm{2log}\:\left(−{x}\right)=\frac{{x}}{\mathrm{25}} \\ $$$$\Rightarrow\frac{\mathrm{ln}\:\left(−{x}\right)}{\mathrm{ln}\:\mathrm{10}}=\frac{{x}}{\mathrm{50}} \\ $$$$\Rightarrow\mathrm{ln}\:\left(−{x}\right)=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}×{x}={ax}\:{with}\:{a}=\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}} \\ $$$$\Rightarrow−{x}={e}^{{ax}} \\ $$$$\Rightarrow\left(−{ax}\right){e}^{−{ax}} ={a} \\ $$$$\Rightarrow−{ax}=\mathbb{W}\left({a}\right) \\ $$$$\Rightarrow{x}=−\frac{{W}\left({a}\right)}{{a}}=−\frac{{W}\left(\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}} \\ $$$$ \\ $$$${all}\:{solutions}: \\ $$$$\Rightarrow{x}=−\frac{{W}\left(\pm\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}\right)}{\frac{\mathrm{ln}\:\mathrm{10}}{\mathrm{50}}}=\begin{cases}{−\frac{{W}\left(\mathrm{0}.\mathrm{046052}\right)}{\mathrm{0}.\mathrm{046052}}=−\frac{\mathrm{0}.\mathrm{044066}}{\mathrm{0}.\mathrm{046052}}=−\mathrm{0}.\mathrm{9569}}\\{−\frac{{W}\left(−\mathrm{0}.\mathrm{046052}\right)}{\mathrm{0}.\mathrm{046052}}=\begin{cases}{−\frac{−\mathrm{0}.\mathrm{048332}}{\mathrm{0}.\mathrm{046052}}=\mathrm{1}.\mathrm{0495}}\\{−\frac{−\mathrm{4}.\mathrm{6052}}{\mathrm{0}.\mathrm{046052}}=\mathrm{100}}\end{cases}}\end{cases} \\ $$

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