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

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. ${\log _b}b = 1$.