Question

If ${\log _x}y,{\log _z}x,{\log _y}z$ are in G.P. $xyz = 64$ and ${x^3},{y^3},{z^3}$ are in A.P then
$A)x = y = z$
$B)x = 4$
$C)x,y,z$ are in G.P
$D)$ All of these

Answer 1

First, We need to know about the concepts of logarithm operations.
We will first understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better which is $\log {x^m} = m\log x$

Formula used:
Using the logarithm law, ${\log _y}x = \dfrac{{\log x}}{{\log y}}$
$\log {x^m} = m\log x$
For A.P the sequence can be expressed as $2b = a + c$ and for G.P ${b^2} = ac$

Complete step-by-step solution:
Since from the given that we have ${\log _x}y,{\log _z}x,{\log _y}z$ are in G.P.
Now convert into the G.P formula we get ${b^2} = ac \Rightarrow {({\log _z}x)^2} = ({\log _x}y.{\log _y}z)$ where a is the first term, b is the second term and c is the third term.
Since ${\log _y}x = \dfrac{{\log x}}{{\log y}}$ then we have ${({\log _z}x)^2} = ({\log _x}y.{\log _y}z) \Rightarrow {(\dfrac{{\log x}}{{\log z}})^2} = (\dfrac{{\operatorname{logy} }}{{\operatorname{logx} }})(\dfrac{{\operatorname{logz} }}{{\log y}})$
Canceling the common terms and cross multiplying we get ${(\dfrac{{\log x}}{{\log z}})^2} = (\dfrac{{\operatorname{logy} }}{{\operatorname{logx} }})(\dfrac{{\operatorname{logz} }}{{\log y}}) \Rightarrow {(logx)^3} = {(logz)^3}$
Thus, taking the cube root and evaluating we have ${(logx)^3} = {(logz)^3} \Rightarrow \log x = \log z \Rightarrow x = z$ where $\log (\dfrac{a}{b}) = \log a - \log b$ which equals to zero.
Thus, we have $x = z$ and take it as the first equation $(1)$
Since from given ${x^3},{y^3},{z^3}$ are in A.P. Now we have $2b = a + c \Rightarrow 2{y^3} = {x^3} + {z^3}$ and we know that $x = z$ then we have $2b = a + c \Rightarrow 2{y^3} = 2{x^3} \Rightarrow y = x$
Therefore, we get $x = y = z$ (option one correct)
Since we have $xyz = 64$ and also, we have $x = y = z$ then we get ${x^3} = 64 \Rightarrow x = 4$ (option two correct)
Also, since ${y^2} = xz \Rightarrow {({\log _z}x)^2} = ({\log _x}y.{\log _y}z)$ (used above)
Thus, we have $x,y,z$ are in G.P (option three correct)
Therefore, all the options are correct, and hence option $D)$ All of these are correct.

Note: While talking about the A.P and G.P, we need to know about the concept of Arithmetic and Geometric progression.
An arithmetic progression can be represented by $a,(a + d),(a + 2d),(a + 3d),...$where $a$ is the first term and $d$ is a common difference.
A geometric progression can be given by $a,ar,a{r^2},....$ where $a$ is the first term and $r$ is a common ratio.