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Question Number 92987 by i jagooll last updated on 10/May/20

{ (((√y) + ln (x^2 )=2)),((y + 4 ln(x) =28)) :}

$$\begin{cases}{\sqrt{\mathrm{y}}\:+\:\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} \right)=\mathrm{2}}\\{\mathrm{y}\:+\:\mathrm{4}\:\mathrm{ln}\left(\mathrm{x}\right)\:=\mathrm{28}}\end{cases} \\ $$

Answered by john santu last updated on 10/May/20

2ln(x) = 14−(y/2) ⇒2ln(x) = 2−(√y) 14−(y/2)=2−(√y) 24−((√y))^2 =−2(√y) ((√y))^2 −2(√y) −24=0 ((√y)−6)((√y)+4)=0 y = 36 ∧ x = e^(−2)

$$\mathrm{2ln}\left(\mathrm{x}\right)\:=\:\mathrm{14}−\frac{\mathrm{y}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{2ln}\left(\mathrm{x}\right)\:=\:\mathrm{2}−\sqrt{\mathrm{y}}\: \\ $$$$\mathrm{14}−\frac{\mathrm{y}}{\mathrm{2}}=\mathrm{2}−\sqrt{\mathrm{y}}\: \\ $$$$\mathrm{24}−\left(\sqrt{\mathrm{y}}\right)^{\mathrm{2}} =−\mathrm{2}\sqrt{\mathrm{y}}\: \\ $$$$\left(\sqrt{\mathrm{y}}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{y}}\:−\mathrm{24}=\mathrm{0} \\ $$$$\left(\sqrt{\mathrm{y}}−\mathrm{6}\right)\left(\sqrt{\mathrm{y}}+\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{y}\:=\:\mathrm{36}\:\wedge\:\mathrm{x}\:=\:\mathrm{e}^{−\mathrm{2}} \: \\ $$

Commented by i jagooll last updated on 10/May/20

thank you

$$\mathrm{thank}\:\mathrm{you} \\ $$

Answered by Rasheed.Sindhi last updated on 10/May/20

Merely another way (Not claiming for better way) (i)⇒y=(2−ln(x^2 ))^2 (ii)⇒y=28−4ln(x) 4−4ln(x^2 )−(ln(x^2 ))^2 =28−4ln(x) 4−8ln(x)−4(ln(x))^2 −28+4ln(x)=0 (ln(x))^2 +ln(x)+7=0 ln(x)=((−1±(√(1−28)))/2) ln(x)=((−1±3i(√3))/2) Can anyone continue?

$${Merely}\:{another}\:{way} \\ $$$$\left({Not}\:{claiming}\:{for}\:{better}\:{way}\right) \\ $$$$\left({i}\right)\Rightarrow{y}=\left(\mathrm{2}−{ln}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{2}} \\ $$$$\left({ii}\right)\Rightarrow{y}=\mathrm{28}−\mathrm{4}{ln}\left({x}\right) \\ $$$$\:\:\:\:\:\:\mathrm{4}−\mathrm{4}{ln}\left({x}^{\mathrm{2}} \right)−\left({ln}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{2}} \:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{28}−\mathrm{4}{ln}\left({x}\right) \\ $$$$\:\:\:\mathrm{4}−\mathrm{8}{ln}\left({x}\right)−\mathrm{4}\left({ln}\left({x}\right)\right)^{\mathrm{2}} −\mathrm{28}+\mathrm{4}{ln}\left({x}\right)=\mathrm{0}\:\:\:\: \\ $$$$\:\:\:\left({ln}\left({x}\right)\right)^{\mathrm{2}} +{ln}\left({x}\right)+\mathrm{7}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:{ln}\left({x}\right)=\frac{−\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{28}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{ln}\left({x}\right)=\frac{−\mathrm{1}\pm\mathrm{3}{i}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:{Can}\:{anyone}\:{continue}? \\ $$

Commented by john santu last updated on 10/May/20

wrong in parts . 4−8ln(x)+4(ln(x))^2 = 28−4ln(x) 4(ln(x))^2 −4ln(x)−24=0 (ln(x))^2 −ln(x)−6=0 {ln(x)−3}{ln(x)+2} = 0 ln(x)= 3 ⇒x= e^3 ln(x)=−2⇒x=e^(−2)

$$\mathrm{wrong}\:\mathrm{in}\:\mathrm{parts}\:. \\ $$$$\mathrm{4}−\mathrm{8ln}\left(\mathrm{x}\right)+\mathrm{4}\left(\mathrm{ln}\left(\mathrm{x}\right)\right)^{\mathrm{2}} \:=\:\mathrm{28}−\mathrm{4ln}\left(\mathrm{x}\right) \\ $$$$\mathrm{4}\left(\mathrm{ln}\left(\mathrm{x}\right)\right)^{\mathrm{2}} −\mathrm{4ln}\left(\mathrm{x}\right)−\mathrm{24}=\mathrm{0} \\ $$$$\left(\mathrm{ln}\left(\mathrm{x}\right)\right)^{\mathrm{2}} −\mathrm{ln}\left(\mathrm{x}\right)−\mathrm{6}=\mathrm{0} \\ $$$$\left\{\mathrm{ln}\left(\mathrm{x}\right)−\mathrm{3}\right\}\left\{\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{2}\right\}\:=\:\mathrm{0} \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)=\:\mathrm{3}\:\Rightarrow\mathrm{x}=\:\mathrm{e}^{\mathrm{3}} \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)=−\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{e}^{−\mathrm{2}} \\ $$$$ \\ $$

Commented by Rasheed.Sindhi last updated on 10/May/20

Thαnk you sir!

$$\mathcal{T}{h}\alpha{nk}\:{you}\:{sir}! \\ $$

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