Question

If \[{x^{18}} = {y^{21}} = {z^{28}}\] , prove that 3, \[3{\log _y}x\] , \[3{\log _z}y\], \[7{\log _x}z\] form an A.P.

Answer 1

Hint:
First we will assume \[3 = a\] , \[3{\log _y}x = b\] , \[3{\log _z}y = c\] and \[7{\log _x}z = d\]. Then we will apply log identity and solve to get the values of a, b, c and d. Then we will write the given equation. Then we will equate this equation to k. Then, we will log both sides. Then, we will use a property of log and solve. Then, we will take the first two elements and find the value of \[\dfrac{{\log x}}{{\log y}}\] , \[\dfrac{{\log y}}{{\log z}}\] and \[\dfrac{{\log z}}{{\log x}}\]. As we get these values and solve, we will get the answer.

Complete step by step solution:
Complete step by step answer:
Let \[a = 3\] ….(1)
and
 \[b = 3lo{g_y}x\]
As \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] , so we get,
 \[ \Rightarrow b = 3\dfrac{{\log x}}{{\log y}}\] ….(2)
Let \[c = 3lo{g_z}y\]
As \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] , so we get,
 \[ \Rightarrow c = 3\dfrac{{\log y}}{{\log z}}\] ....(3)
Let \[d = 7lo{g_x}z\]
As \[{\log _b}a = \dfrac{{\log a}}{{\log b}}\] , so we get,
 \[ \Rightarrow b = 7 \times \dfrac{{\log z}}{{\log x}}\] ….(4)
To find the value of a, b, c and d, first we will find the value of \[\dfrac{{\log x}}{{\log y}}\] , \[\dfrac{{\log y}}{{\log z}}\] and \[\dfrac{{\log z}}{{\log x}}\] .
Given that \[{x^{18}} = {y^{21}} = {z^{28}}\]
Suppose \[{x^{18}} = {y^{21}} = {z^{28}} = k\] ….(5)
Taking log
 \[ \Rightarrow \log {x^{18}} = \log {y^{21}} = \log {z^{28}} = \log k\] ….(6)
In equation (2), we use the following property, \[\log {a^b} = b\log a\]
 \[ \Rightarrow 18\log x = 21\log y = 28\log z = \log k\] ….(7)
Taking first two terms of equation (7), we have
 \[ \Rightarrow 18\log x = 21\log y\]
 \[ \Rightarrow \dfrac{{\log x}}{{\log y}} = \dfrac{{21}}{{18}}\] ….(8)
Similarly, taking middle two terms of equation (7)
 \[ \Rightarrow 21\log y = 28\log z\]
On simplification we get,
 \[ \Rightarrow \dfrac{{\log y}}{{\log z}} = \dfrac{{28}}{{21}}\] ….(9)
Taking first and third terms of equation (7)
 \[ \Rightarrow 18\log x = 28\log z\]
On simplification we get,
 \[ \Rightarrow \dfrac{{\log z}}{{\log x}} = \dfrac{{18}}{{28}}\] ….(10)
From equation (1)
  \[ \Rightarrow a = 3\]
And from equation (2)
 \[ \Rightarrow b = 3\dfrac{{\log x}}{{\log y}}\]
Here, we will put the value of equation (8)
Therefore,
 \[ \Rightarrow b = 3 \times \dfrac{{21}}{{18}}\]
On simplification we get,
 \[ \Rightarrow b = \dfrac{7}{2}\]
 \[\therefore b = \dfrac{7}{2}\]
Now, we will determine the value of c
From equation (3), we have
 \[ \Rightarrow c = 3\dfrac{{\log y}}{{\log z}}\]
Putting the values of equation (9)
 \[ \Rightarrow c = 3 \times \dfrac{{28}}{{21}}\]
 \[\therefore c = 4\]
From equation (4)
 \[ \Rightarrow d = \times \dfrac{{\log z}}{{\log x}}\]
Put the value of equation (10) in equation (4), we get
 \[ \Rightarrow d = 7 \times \dfrac{{18}}{{28}}\]
On simplification, we get
 \[ \Rightarrow d = \dfrac{9}{2}\]
Therefore, the values of a, b, c and d are
 \[a = 3,b = \dfrac{7}{2},c = 4,d = \dfrac{9}{2}\]
i.e.
 \[a = 3,b = 3.5,c = 4,d = 4.5\]
The common difference is 0.5.

Hence, these terms combine to form an A.P.

Note:
You may find trouble in applying log identities and solving them. Logarithms are the inverse of exponentials. Logarithm with base 10 of 10 i.e. \[{\log _{10}}10 = 1\] . Logarithm with base 10 of 100 is 2 i.e. \[{\log _{10}}100 = 2\] . And the logarithm with base 10 of 1000 is 3 i.e. \[{\log _{10}}1000 = 3\] . Logarithm of some value with the same base value is always 1 i.e. \[{\log _b}b = 1\].