Question

plaese llok at the attacherment and solve the questions palaeser

Answer 1

Given      x, y, z > 0.          x ≠ 1  ,  y ≠ 1    z ≠  1

$\frac{Log\ x}{y-z}=\frac{log\ y}{z-x}=\frac{log\ z}{x-y}=k\ (say)\\\\Log\ x=k(y-z)\\Log\ y=k(z-x)\\Log\ z=k(x-y)\\\\Log\ x^x\ y^y\ z^z=x\ Log\ x+y\ Log\ y+z\ log\ z\\.\ \ \ =x*k(y-z)+y*k(z-x)+z*k(x-y)\\.\ \ \ =k(xy-xz+yz-yx+zx-zy)=0\\\\x^x\ y^y\ z^z=1\\\\Also,\\.\ \ \ Log\ x+log\ y+log\ z=0\ \ \ =>\ Log\ xyz=0\ \ =>xyz=1$
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$Log_2\ x\ *Log_2\ xyz=48,\ \ Log_2\ y\ *Log_2\ xyz=12,\\Log_2\ z\ *Log_2\ xyz=84\\\\Let\ x=2^a,\ \ y=2^b,\ \ \ z=2^c\\.\ \ \ =>Log_2x=a,\ \ Log_2y=b,\ \ Log_2z=c\\\\Also,\ \ xyz=2^{a+b+c},\ \ Log_2\ xyz=a+b+c\\\\Given:\\.\ \ \ \ a(a+b+c)=48,\ \ \ b(a+b+c)=12,\ \ \ c(a+b+c)=84\\\\Add\ all\ the\ three\ equations\\.\ \ \ \ (a+b+c)(a+b+c)=48+12+84=144\\.\ \ \ \ a+b+c=+12\ \ or\ \ -12\\\\Hence,\\\\a=4,\ \ b=1,\ \ c=7,\ \ =>\ \ \ x=2^4=16,\ \ y=2,\ \ z=2^7=128$

OR

$a=-4,\ \ b=-1,\ \ c=-7,\ \ \ \ =>x=\frac{1}{2^4},\ \ y=\frac{1}{2},\ \ z=\frac{1}{2^7}$
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$X=8\ Log_2\ ( \sqrt[3]{121}+\frac{1}{3} )\\\\=Log_2\ (11^{\frac{2}{3}}+\frac{1}{3})^8$

is the problem correctly mentioned ?  I dont see  logical solution in this, other than using a calculator.