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Question: If \[{x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > {1_1}\] then the value of \[n{\text{ }}log{\text{ }...

If xn>xn−1>...>x2>x1>11{x_n} > {x_{n - 1}} > ... > {x_2} > {x_1} > {1_1} then the value of n log mn{\text{ }}log{\text{ }}m
A.00
B.11
C.22

Explanation

Solution

In the given question, we are given that xn>xn−1....>x2>x1>1{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 xn,xn−1{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⁡x1log⁡x2log⁡x3....log⁡xnxn(xn−1)....x1{\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 xn,xn−1,.....,x2,x1,1{x_n},{x_{n - 1}},.....,{x_2},{x_1},1
Are given in an order which isxn,>xn−1>.....>x2>x1>1{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⁡x1log⁡x2log⁡x3....log⁡xnxn(xn−1)....x1{\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⁡mn=nlog⁡m\log {m^n} = n\log m
We get log⁡x1log⁡x2log⁡x3....log⁡xnxn(xn−1)....x1{\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}....{\log _{{x_n}}}{x_n}^{\left( {{x_{n - 1}}} \right){{....}^{{x_1}}}} x1log⁡xnxn{x_{{1_{{{\log }_{^{^{{x_n}^{{x_n}}}}}}}}}}
Also log⁡mm=1{\log _m}m = 1 therefore log⁡xnxn=1{\log _{{x_n}}}{x_n} = 1
We get log⁡x1log⁡x2log⁡x3....log⁡xnxn(xn−1)....x1{\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⁡x1log⁡x2log⁡x3x3x2x1...............(1){\log _{{x_1}}}{\log _{{x_2}}}{\log _{{x_3}}}{x_3}^{{x_2}^{{x_1}}}...............(1)
Again, in the third term log⁡x3x3(x2x1){\log _{{x_3}}}{x_3}^{\left( {{x_2}^{{x_1}}} \right)}
We get x2x1log⁡x3x3{x_2}^{{x_1}}{\log _{{x_3}}}{x_3}
Since log⁡mm=1{\log _m}m = 1therefore log⁡x3x3=1{\log _{{x_3}}}{x_3} = 1
Therefore equation 11 becomes log⁡x1log⁡x2x2x1{\log _{{x_1}}}{\log _{{x_2}}}{x_2}^{{x_1}}
Again using the same logarithmic identities, we get
log⁡x1,x1log⁡x2x2{\log _{{x_1}}},{x_1}{\log _{{x_2}^{{x_2}}}}
Again log⁡x2x2=1{\log _{{x^2}}}{x_2} = 1, we get log⁡x1x1{\log _{{x_1}}}{x_1}
Which is log⁡x1log⁡x2log⁡x3....log⁡xnxn(xn−1)....x1=1{\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⁡mn=1{\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⁡mm=1{\log _{{m^m}}} = 1. Also, second identity is log⁡mn\log {m^n}, which means log of m to the power n, then the value of log⁡mn\log {m^n} becomes n logmn{\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⁡mn=log⁡m−log⁡n\log mn = \log m + \log n,\log \dfrac{m}{n} = \log m - \log n
log⁡mn=nlog⁡m\log {m^n} = n\log m and log⁡mm=1{\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.