Question

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.\[2\]

Answer 1

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.

Complete step-by-step answer:
 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.