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Question-121100




Question Number 121100 by Ar Brandon last updated on 05/Nov/20
Answered by TANMAY PANACEA last updated on 05/Nov/20
((logx)/(x(y+z−x)))=((logy)/(y(z+x−y)))=((logz)/(z(x+y−z)))=k  logx=kx(y+z−x)  logy=ky(z+x−y)  logz=kz(x+y−z)  A)x^y y^x =p  ylogx+xlogy=logp  ykx(y+z−x)+kxy(z+x−y)=logp  xyk(y+z−x+z+x−y)  xyk(2z)=2xyz×k  y^z z^y =Q  zlogy+ylogz=logQ  zky(z+x−y)+ykz(x+y−z)=logQ  yzk(2x)  2xyzk  so logp=logQ  so option A is correct...  pls check same way option (B,C,D) pls
$$\frac{{logx}}{{x}\left({y}+{z}−{x}\right)}=\frac{{logy}}{{y}\left({z}+{x}−{y}\right)}=\frac{{logz}}{{z}\left({x}+{y}−{z}\right)}={k} \\ $$$${logx}={kx}\left({y}+{z}−{x}\right) \\ $$$${logy}={ky}\left({z}+{x}−{y}\right) \\ $$$${logz}={kz}\left({x}+{y}−{z}\right) \\ $$$$\left.{A}\right){x}^{{y}} {y}^{{x}} ={p} \\ $$$${ylogx}+{xlogy}={logp} \\ $$$${ykx}\left({y}+{z}−{x}\right)+{kxy}\left({z}+{x}−{y}\right)={logp} \\ $$$${xyk}\left({y}+{z}−{x}+{z}+{x}−{y}\right) \\ $$$${xyk}\left(\mathrm{2}{z}\right)=\mathrm{2}{xyz}×{k} \\ $$$${y}^{{z}} {z}^{{y}} ={Q} \\ $$$${zlogy}+{ylogz}={logQ} \\ $$$${zky}\left({z}+{x}−{y}\right)+{ykz}\left({x}+{y}−{z}\right)={logQ} \\ $$$${yzk}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{2}{xyzk} \\ $$$${so}\:{logp}={logQ} \\ $$$${so}\:{option}\:{A}\:{is}\:{correct}… \\ $$$${pls}\:{check}\:{same}\:{way}\:{option}\:\left({B},{C},{D}\right)\:{pls} \\ $$$$ \\ $$
Commented by Ar Brandon last updated on 05/Nov/20
Thank you Sir. It is a multiple correct choice type question.
Commented by TANMAY PANACEA last updated on 05/Nov/20
most welcome sir
$${most}\:{welcome}\:{sir} \\ $$

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