Question

then xy.z? is:
logx logy logz
If
y-Z
Z-X X-y
(A) x + y + z
(C) - 1
(B) O
(D) 1​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\blue{\dfrac{log(x)}{y-z}=\dfrac{log(y)}{z-x}=\dfrac{log(z)}{x-y}}}$

Now, let

$\sf{\dfrac{log(x)}{y-z}=\dfrac{log(y)}{z-x}=\dfrac{log(z)}{x-y}=k}$

So,

$\sf{log(x)=k(y-z)\,\,\,\,,log(y)=k(z-x)\,\,\,\,,log(z)=k(x-y)}$

Now,

$\sf{\purple{Let\,\,\,\eta=x^{x}\cdot\,y^{y}\cdot\,z^{z}}}$

$\sf{\implies\,log(\eta)=log(x^{x}\cdot\,y^{y}\cdot\,z^{z})}$

$\sf{\implies\,log(\eta)=log(x^{x})+log(y^{y})+log(z^{z})}$

$\sf{\implies\,log(\eta)=x\,log(x)+y\,log(y)+z\,log(z)}$

$\sf{\implies\,log(\eta)=x\cdot\,k(y-z)+y\cdot\,k(z-x)+z\cdot\,k(x-y)}$

$\sf{\implies\,log(\eta)=k\{x(y-z)+y(z-x)+z(x-y)\}}$

$\sf{\implies\,log(\eta)=k\{xy-zx+yz-xy+zx-yz\}}$

$\sf{\implies\,log(\eta)=k\{0\}}$

$\sf{\implies\,log(\eta)=0}$

$\sf{\implies\,\eta=e^0}$

$\sf{\implies\,\eta=1}$

$\bf{\green{Hence,\,\,\,x^{x}\cdot\,y^{y}\cdot\,z^{z}=1}}$