If x^ ((y)/(z)) · y^ (X z)^2 · z^ ((x)/(y)) =y^ 4 y for any x > 1, y > 1 and z > 1 , then which one of the following is correct?

Correct answer: B. n y is the A M of nx, nx, nx and n z x^ ((y)/(z)) · y^ (x z)^2 · z^ ((x)/(y)) =y^ 4 y aligned & [x^ ((y)/(z)) ]+ [y^ (x z)^2 ]+ [z^ ((x)/(y)) ]= [y^ 4 y ] & [ ((y)/(z)) x]+[2 (x z) y]+[ ((x)/(y)) z]=4[ y]^2 & x[ y- z]+2 y[ x+ z] & + z[ x- y]=4[ y]^2 aligned aligned & 3 x+ z=4 y & x+ x+ x+ z 4 = y aligned y is the A M of x, x, x, x \ and z .

NDA 2016

Exam: NDA
Year: 2016
Subject: Mathematics
Chapter: Basics of Mathematics

NDA 2016 · Mathematics · Basics of Mathematics

If x^ ((y)/(z)) · y^ (X z)^2 · z^ ((x)/(y)) =y^ 4 y for any x > 1,...

x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (X z)^2} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 \ln y}

for any

x > 1, y > 1

and

z > 1

, then which one of the following is correct?

Options

A $\ell \mathrm{n} \mathrm{y}$ is the GM of $\ell \mathrm{nx}, \ell \mathrm{nx}, \ell \mathrm{nx}$ and $\ell \mathrm{nz}$
B $\ell \mathrm{n} y$ is the $A M$ of $\ell \mathrm{nx}, \ell \mathrm{nx}, \ell \mathrm{nx}$ and $\ell \mathrm{n} \mathrm{z}$
C $\ell \mathrm{n} \mathrm{y}$ is the HM of $\ell \mathrm{nx}, \ell \mathrm{n} \mathrm{x}$ and $\ell \mathrm{nz}$
D $\ell \mathrm{n} y$ is the AM of $\ell \mathrm{n}, \operatorname{In} \mathrm{x}, \ell \mathrm{n} \mathrm{z}$ and $\ell \mathrm{n} \mathrm{z}$

Correct answer

B. $\ell \mathrm{n} y$ is the $A M$ of $\ell \mathrm{nx}, \ell \mathrm{nx}, \ell \mathrm{nx}$ and $\ell \mathrm{n} \mathrm{z}$

Step-by-step solution

x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (x z)^2} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 \ln y}

\begin{aligned} & \Rightarrow \ln \left[x^{\ln \left(\frac{y}{z}\right)}\right]+\ln \left[y^{\ln (x z)^2}\right]+\ln \left[z^{\ln \left(\frac{x}{y}\right)}\right]=\ln \left[y^{4 \ln y}\right] \\ & \Rightarrow\left[\ln \left(\frac{y}{z}\right) \ln x\right]+[2 \ln (x z) \ln y]+\left[\ln \left(\frac{x}{y}\right) \ln z\right]=4[\ln y]^2 \\ & \Rightarrow \ln \mathrm{x}[\ln \mathrm{y}-\ln \mathrm{z}]+2 \ln \mathrm{y}[\ln \mathrm{x}+\ln \mathrm{z}] \\ & +\ln \mathrm{z}[\ln \mathrm{x}-\ln \mathrm{y}]=4[\ln \mathrm{y}]^2 \end{aligned}

$\begin{aligned} & \Rightarrow 3 \ln x+\ln \mathrm{z}=4 \ln \mathrm{y} \\ & \Rightarrow \frac{\ln \mathrm{x}+\ln \mathrm{x}+\ln \mathrm{x}+\ln \mathrm{z}}{4}=\ln \mathrm{y} \end{aligned}

\therefore \quad \ln y

is the

A M of \ln x, \ln x, \ln x, \ln x \ \text{ and } \ln z$.

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