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Question Number 42289 by mondodotto@gmail.com last updated on 22/Aug/18
log(x+y)=1 log_2 x+log_4 y^2 =4
$$\boldsymbol{\mathrm{log}}\left(\boldsymbol{{x}}+\boldsymbol{{y}}\right)=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{log}}_{\mathrm{2}} \boldsymbol{{x}}+\boldsymbol{\mathrm{log}}_{\mathrm{4}} \boldsymbol{{y}}^{\mathrm{2}} =\mathrm{4} \\ $$
Commented by math khazana by abdo last updated on 22/Aug/18
⇒x+y =e and ((ln(x))/(ln(2))) +((ln(y^2 ))/(ln(4))) =4 ⇒ x+y =e and ln(x) +ln(y) =4ln(2) =ln(2^4 ) ⇒ x+y =e and ln(xy) =ln(16) ⇒ x+y =e and xy =16 ⇒ xand y are solution of the equation u^2 −eu +16 =0 Δ =e^2 −4(16) =e^2 −64 if x and y are real ⇒no solution if x and y are complex ⇒ u_1 =((e+i(√(64−e^2 )))/2) u_2 =((e −i(√(64−e^2 )))/2) .
$$\Rightarrow{x}+{y}\:={e}\:{and}\:\:\frac{{ln}\left({x}\right)}{{ln}\left(\mathrm{2}\right)}\:+\frac{{ln}\left({y}^{\mathrm{2}} \right)}{{ln}\left(\mathrm{4}\right)}\:=\mathrm{4}\:\Rightarrow \\ $$$${x}+{y}\:={e}\:{and}\:{ln}\left({x}\right)\:+{ln}\left({y}\right)\:=\mathrm{4}{ln}\left(\mathrm{2}\right)\:={ln}\left(\mathrm{2}^{\mathrm{4}} \right)\:\Rightarrow \\ $$$${x}+{y}\:={e}\:{and}\:{ln}\left({xy}\right)\:={ln}\left(\mathrm{16}\right)\:\Rightarrow \\ $$$${x}+{y}\:={e}\:{and}\:{xy}\:=\mathrm{16}\:\Rightarrow\:{xand}\:{y}\:{are}\:{solution}\:{of}\:{the} \\ $$$${equation}\:{u}^{\mathrm{2}} \:−{eu}\:+\mathrm{16}\:=\mathrm{0} \\ $$$$\Delta\:={e}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{16}\right)\:={e}^{\mathrm{2}} −\mathrm{64} \\ $$$${if}\:{x}\:{and}\:{y}\:{are}\:{real}\:\Rightarrow{no}\:{solution}\:{if}\:{x}\:{and}\:{y}\:{are} \\ $$$${complex}\:\Rightarrow\:{u}_{\mathrm{1}} =\frac{{e}+{i}\sqrt{\mathrm{64}−{e}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${u}_{\mathrm{2}} =\frac{{e}\:−{i}\sqrt{\mathrm{64}−{e}^{\mathrm{2}} }}{\mathrm{2}}\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/Aug/18
x+y=e^1 =e ((lnx)/(ln2))+((2lny)/(2ln2))=4 lnx+lny=4ln2 ln(xy)=ln2^4 xy=16 x+y=e x(e−x)=16 xe−x^2 −16=0 x^2 −xe+16=0 x=((e±(√(e^2 −64)))/2)=((e±i(√(64−e^2 )))/2) y=e−((e±(√(e^2 −64)))/2)=((e−_+ (√(e^2 −64)))/2)=((e−_+ i(√(64−e^2 )))/2) or if log(x+y)=1 log_(10) (x+y)=1 x+y=10 then x=((10±(√(100−64)))/2)=((10±6)/2)=8 and 2 so y=2 and 8
$${x}+{y}={e}^{\mathrm{1}} ={e} \\ $$$$\frac{{lnx}}{{ln}\mathrm{2}}+\frac{\mathrm{2}{lny}}{\mathrm{2}{ln}\mathrm{2}}=\mathrm{4} \\ $$$${lnx}+{lny}=\mathrm{4}{ln}\mathrm{2} \\ $$$${ln}\left({xy}\right)={ln}\mathrm{2}^{\mathrm{4}} \\ $$$${xy}=\mathrm{16} \\ $$$${x}+{y}={e} \\ $$$${x}\left({e}−{x}\right)=\mathrm{16} \\ $$$${xe}−{x}^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$${x}^{\mathrm{2}} −{xe}+\mathrm{16}=\mathrm{0} \\ $$$${x}=\frac{{e}\pm\sqrt{{e}^{\mathrm{2}} −\mathrm{64}}}{\mathrm{2}}=\frac{{e}\pm{i}\sqrt{\mathrm{64}−{e}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$${y}={e}−\frac{{e}\pm\sqrt{{e}^{\mathrm{2}} −\mathrm{64}}}{\mathrm{2}}=\frac{{e}\underset{+} {−}\sqrt{{e}^{\mathrm{2}} −\mathrm{64}}}{\mathrm{2}}=\frac{{e}\underset{+} {−}{i}\sqrt{\mathrm{64}−{e}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\boldsymbol{{or}} \\ $$$${if}\:{log}\left({x}+{y}\right)=\mathrm{1}\:\:\:\:\:\:{log}_{\mathrm{10}} \left({x}+{y}\right)=\mathrm{1} \\ $$$${x}+{y}=\mathrm{10} \\ $$$${then}\:{x}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{64}}}{\mathrm{2}}=\frac{\mathrm{10}\pm\mathrm{6}}{\mathrm{2}}=\mathrm{8}\:\:{and}\:\mathrm{2} \\ $$$${so}\:{y}=\mathrm{2}\:{and}\:\mathrm{8} \\ $$

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