Question

If ${\log }_x\left(y\right),{\log }_z\left(x\right),{\log }_y\left(z\right)$ are in G.P., $xyz=64$ and $x^3,y^3,z^3$ are in A.P., then:

• A. $x=y=z=4$
• B. $x=4$
• C. $x,y,z$ are in G.P.
• D. All of these

Answer 1

The correct option is D

All of these

Explanation for the correct option:

It is given that ${\log }_x\left(y\right),{\log }_z\left(x\right),{\log }_y\left(z\right)$ are in G.P.

We know that if $a,b,c$ are in G.P. then $b^2=ac$

$$\begin{gathered} \Rightarrow {\left({\log }_z\left(x\right)\right)}^2={\log }_x\left(y\right)·{\log }_y\left(z\right) \\ \Rightarrow {\left(\frac{\log \left(x\right)}{\log \left(z\right)}\right)}^2=\frac{\cancel{\log \left(y\right)}}{\log \left(x\right)}·\frac{\log \left(z\right)}{\cancel{\log \left(y\right)}}\mathbf{[}\mathbf{\because }\mathit{l}\mathit{o}{\mathit{g}}_{\mathbf{a}}\left(b\right)\mathbf{=}\frac{\mathbf{l}\mathbf{o}\mathbf{g}\left(b\right)}{\mathbf{l}\mathbf{o}\mathbf{g}\left(a\right)}] \\ \Rightarrow {\left(\log \left(x\right)\right)}^3={\left(\log \left(z\right)\right)}^3 \\ \Rightarrow x=z \end{gathered}$$

Now, it is also given that $x^3,y^3,z^3$ are in A.P.

We know that if $a,b,c$ are in A.P. then $2b=a+c$

So, $2y^3=x^3+z^3$

$$\begin{gathered} \Rightarrow 2y^3=z^3+z^3\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{x}\mathbf{=}\mathbf{z}\mathbf{]}\mathbf{} \\ \Rightarrow 2y^3=2z^3 \\ \Rightarrow y=z \end{gathered}$$

Thus, $x=y=z$, satisfying option (A)

Also, given that $xyz=64$

So, $x^3=64$

Hence, $x=4$ and $x=y=z=4$, satisfying option (B)

Here, $y^2=xz$, which is a condition for G.P, satisfying option (C)

Hence, option(D) i.e. All of these is the correct answer.