1708Yes Rahul it is (log(z))^2
\hspace{-16}$If $\bf{xyz=10^{81}}$ and $\bf{\ln(x).\ln(yz)+\ln(y).\ln(z)=468}$\\\\\\ Then $\bf{\sqrt{(\ln(x))^2+(\ln(y))^2+\ln(z)^2}=}$\\\\\\ Where $\bf{x,y,z\in \mathbb{R^{+}}}$
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3 Answers
36i think there's a typographical mistake in ur question...!!
maybe its log in place of ln and its (ln(z))2 in the part completely inside the square root...!!
1logx+logy+logz=81
logx.log(yz)+logy.logz=468
=>logx.logy+logy.logz+logx.logz=468
take a=logx,b=logy,c=logz
so
a+b+c=81
ab+bc+ac=468
(a+b+c)2=a2+b2+c2+2(ab+bc+ac)
a2+b2+c2=(a+b+c)2-2(ab+bc+ac)=812-2X468=5625
therefore,
√a2+b2+c2=√5625=75
i have taken log instead of ln