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If $x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (x z)^{2}} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 / n y}$ for any $x>1, y>1$ and $z>1$,
then which one of the following is correct?
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then which one of the following is correct?
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Verified Answer
The correct answer is:
$\ell \mathrm{n} \mathrm{y}$ is the $\mathrm{AM}$ of $\ell \mathrm{n} \mathrm{x}, \ell \mathrm{n} \mathrm{x}, \ln \mathrm{x}$ and $\ell \mathrm{n} \mathrm{z}$
$x^{\ln \left(\frac{y}{z}\right)} \cdot y^{\ln (x z)^{2}} \cdot z^{\ln \left(\frac{x}{y}\right)}=y^{4 \ln y}$
$\Rightarrow \ln \left[x^{ \left.\ln \left(\frac{y}{z}\right)\right]}+\ln \left[y^{\ln (x z)^{2}}\right]+\ln \left[z^{ \left.\ln \left(\frac{x}{y}\right)\right]=\ln \left[y^{4 \ln y}\right]}\right.\right.$
$\Rightarrow\left[\ln \left(\frac{y}{z}\right) \ln x\right]+[2 \ln (x z) \ln y]+\left[\ln \left(\frac{x}{y}\right) \ln z\right]=4[\ln y]^{2}$
$\Rightarrow \ln x[\ln y-\ln z]+2 \ln y[\ln x+\ln z]$
$\quad+\ln z[\ln x-\ln y]=4[\ln y]^{2}$
$\Rightarrow 3 \ln x+\ln z=4 \ln y$
$\Rightarrow \frac{\ln x+\ln x+\ln x+\ln z}{4}=\ln y$
$\therefore \quad$ lny is the AM of $\ln x, \ln x, \ln x, \ln x \& \ln z .$
$\Rightarrow \ln \left[x^{ \left.\ln \left(\frac{y}{z}\right)\right]}+\ln \left[y^{\ln (x z)^{2}}\right]+\ln \left[z^{ \left.\ln \left(\frac{x}{y}\right)\right]=\ln \left[y^{4 \ln y}\right]}\right.\right.$
$\Rightarrow\left[\ln \left(\frac{y}{z}\right) \ln x\right]+[2 \ln (x z) \ln y]+\left[\ln \left(\frac{x}{y}\right) \ln z\right]=4[\ln y]^{2}$
$\Rightarrow \ln x[\ln y-\ln z]+2 \ln y[\ln x+\ln z]$
$\quad+\ln z[\ln x-\ln y]=4[\ln y]^{2}$
$\Rightarrow 3 \ln x+\ln z=4 \ln y$
$\Rightarrow \frac{\ln x+\ln x+\ln x+\ln z}{4}=\ln y$
$\therefore \quad$ lny is the AM of $\ln x, \ln x, \ln x, \ln x \& \ln z .$
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