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# If ${x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > {1_1}$ then the value of $n{\text{ }}log{\text{ }}m$A.$0$B.$1$C.$2$

Last updated date: 20th Jun 2024
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Hint:In the given question, we are given that ${x_n} > {x_{n - 1}}.... > {x_2} > {x_1} > 1$ where n is any constant and further we have to find out the value of these ${x_n},{x_{n - 1}}$ with logarithm. Using logarithmic identities, We will find the value of these logarithm values asked in the question.

In the given question we have to find out the value of
${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}}$
Where the values of ${x_n},{x_{n - 1}},.....,{x_2},{x_1},1$
Are given in an order which is${x_{n,}} > {x_{n - 1}} > ..... > {x_2} > {x_1} > 1$
Since we have to find the value in log.
Therefore, using logarithmic identities, we will find the value of asked question. We are to find the value of
${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}}$
Now using the identity $\log {m^n} = n\log m$
We get ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}}$ ${x_{{1_{{{\log }_{^{^{{x_n}^{{x_n}}}}}}}}}}$
Also ${\log _m}m = 1$ therefore ${\log _{{x_n}}}{x_n} = 1$
We get ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}}$
Therefore, we get ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}{x_3}^{{x_2}^{{x_1}}}...............(1)$
Again, in the third term ${\log _{{x_3}}}{x_3}^{\left( {{x_2}^{{x_1}}} \right)}$
We get ${x_2}^{{x_1}}{\log _{{x_3}}}{x_3}$
Since ${\log _m}m = 1$therefore ${\log _{{x_3}}}{x_3} = 1$
Therefore equation $1$ becomes ${\log _{{x_1}}}{\log _{{x_2}}}{x_2}^{{x_1}}$
Again using the same logarithmic identities, we get
${\log _{{x_1}}},{x_1}{\log _{{x_2}^{{x_2}}}}$
Again ${\log _{{x^2}}}{x_2} = 1$, we get ${\log _{{x_1}}}{x_1}$
Which is ${\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}} = 1$
So option (B) is correct.
In this question, We had used two identities, that for two constants or variables m and n ${\log _{{m^n}}} = 1$ which means if both in the base and in the value if there is same constant or variable then the value of ${\log _{{m^m}}} = 1$. Also, second identity is $\log {m^n}$, which means log of m to the power n, then the value of $\log {m^n}$ becomes $n{\text{ }}logm$ that means power comes in front of the log and the value of m is there.

Note: Logarithmic is the mathematical expression or formula that is used to find out the values of various variables and constants. Some basic identities of logarithm for two variables m and n are $\log mn = \log m + \log n,\log \dfrac{m}{n} = \log m - \log n$
$\log {m^n} = n\log m$ and ${\log _{{m^m}}} = 1$
Using these logarithmic identities, we can find out the values of variables or constants as well as the relationship between them.