Answer
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Hint: We are given that \[{x^{18}} = {y^{21}} = {z^{28}}\]. We will assume a variable m is equal to this equation and then we will apply logarithm to both the sides. After that, we will assume the given terms which are required to be proved in an A. P as some random variables and we will calculate their individual values first. Once we have calculated the value of the terms given, we can prove that they are in A. P by showing that their common difference d is equal.
Complete step-by-step answer:
We are given an equation that: \[{x^{18}} = {y^{21}} = {z^{28}}\].
Let us assume that m is equal to this equation i. e.,
\[{x^{18}} = {y^{21}} = {z^{28}}\]= $m$
Now, taking logarithms both the sides, we get
$\log {x^{18}} = \log {y^{21}} = \log {z^{28}} = \log m$
We know that there is a logarithmic identity which states $\log {a^b} = b\log a$. Using this property in the above equation, we can write it as
$18\log x = 21\log y = 28\log z = \log m$
Now, considering $18\log x = 21\log y$, we can write it as
$\dfrac{{\log x}}{{\log y}} = \dfrac{{21}}{{18}}$
Similarly, $\dfrac{{\log y}}{{\log z}} = \dfrac{{28}}{{21}}$
And, $\dfrac{{\log z}}{{\log x}} = \dfrac{{18}}{{28}}$
Now, considering the series given: $3,3{\log _y}x,3{\log _z}y,7{\log _x}z$
Let us assume a = 3, b =3${\log _y}x$ , c =3${\log _z}y$ and d = 7${\log _x}z$
Now, we will simplify these terms and we will use the values we have obtained above to get their values from which we will determine if the given terms combine to form an A.P. or not.
Considering b = 3${\log _y}x$
We can say b = $3\dfrac{{\log x}}{{\log y}}$ = $3 \times \dfrac{{21}}{{18}} = \dfrac{7}{2}$= 3.5
Similarly, c = $3\dfrac{{\log y}}{{\log z}}$= $3 \times \dfrac{{28}}{{21}} = 4$
And, d = $7\dfrac{{\log z}}{{\log x}}$= $7 \times \dfrac{{18}}{{28}} = 4.5$
Now, we have the values as: a = 3, b = 3.5, c = 4 and d = 4.5
We can tell even by observation that their common difference is equal i.e., d = 0.5
Hence, it is proved that these terms combine to form an A.P.
Note: In such problems, you may face difficulties while simplifying the given equations to calculate their values such as when we used the logarithmic property or when we converted the logarithmic function to the rational form. Once you convert the equation then it is easy to solve.
Complete step-by-step answer:
We are given an equation that: \[{x^{18}} = {y^{21}} = {z^{28}}\].
Let us assume that m is equal to this equation i. e.,
\[{x^{18}} = {y^{21}} = {z^{28}}\]= $m$
Now, taking logarithms both the sides, we get
$\log {x^{18}} = \log {y^{21}} = \log {z^{28}} = \log m$
We know that there is a logarithmic identity which states $\log {a^b} = b\log a$. Using this property in the above equation, we can write it as
$18\log x = 21\log y = 28\log z = \log m$
Now, considering $18\log x = 21\log y$, we can write it as
$\dfrac{{\log x}}{{\log y}} = \dfrac{{21}}{{18}}$
Similarly, $\dfrac{{\log y}}{{\log z}} = \dfrac{{28}}{{21}}$
And, $\dfrac{{\log z}}{{\log x}} = \dfrac{{18}}{{28}}$
Now, considering the series given: $3,3{\log _y}x,3{\log _z}y,7{\log _x}z$
Let us assume a = 3, b =3${\log _y}x$ , c =3${\log _z}y$ and d = 7${\log _x}z$
Now, we will simplify these terms and we will use the values we have obtained above to get their values from which we will determine if the given terms combine to form an A.P. or not.
Considering b = 3${\log _y}x$
We can say b = $3\dfrac{{\log x}}{{\log y}}$ = $3 \times \dfrac{{21}}{{18}} = \dfrac{7}{2}$= 3.5
Similarly, c = $3\dfrac{{\log y}}{{\log z}}$= $3 \times \dfrac{{28}}{{21}} = 4$
And, d = $7\dfrac{{\log z}}{{\log x}}$= $7 \times \dfrac{{18}}{{28}} = 4.5$
Now, we have the values as: a = 3, b = 3.5, c = 4 and d = 4.5
We can tell even by observation that their common difference is equal i.e., d = 0.5
Hence, it is proved that these terms combine to form an A.P.
Note: In such problems, you may face difficulties while simplifying the given equations to calculate their values such as when we used the logarithmic property or when we converted the logarithmic function to the rational form. Once you convert the equation then it is easy to solve.
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