Question

The expression \[{\mathbf{lo}}{{\mathbf{g}}_{\mathbf{p}}}\]\[{\log _p}\sqrt[p]{{\sqrt[p]{{\sqrt[p]{{............\sqrt[p]{p}}}}}}}\], where \[p \geqslant 2,\]\[p \in N\]\[;n \in N\] when simplified is
A. Independent of p
B. Independent of p and of n
C. Dependent on both p and n
D. Positive

Answer 1

Hint: In this we do not simplify the whole question while we simplify first the radical term I,e.
The term which contain ‘p’ terms simplified as
\[\sqrt[p]{{\sqrt[p]{{\sqrt[p]{{.........\sqrt[p]{p}}}}}}}\] (n times radical sign)
\[ = {\left( {{{\left( {{{\left( {{{\left( p \right)}^{\dfrac{1}{p}}}} \right)}^{\dfrac{1}{p}}}} \right)}^{\dfrac{1}{p}}}} \right)^{\dfrac{1}{p}}}\](n times)
According to the property:
\[{\left( {{x^a}} \right)^b}\, = \,{x^{ab}}\]

Complete step by step solution:
Simplifying from inside
\[ \Rightarrow {\log _p}\,{\log _p}\sqrt[p]{{\sqrt[p]{{\sqrt[p]{{......\sqrt[p]{p}}}}}}}\]
\[ \Rightarrow {\log _p}\,{\log _p}{\left( {{{\left( {{{\left( {{{\left( {{{\left( p \right)}^{\dfrac{1}{p}}}} \right)}^{^{\dfrac{1}{p}}}}} \right)}^{^{\dfrac{1}{p}}}}} \right)}^{........}}} \right)^{\dfrac{1}{p}}}\](n times)
\[ \Rightarrow {\log _p}\,{\log _p}\,\left( {{{\left( p \right)}^{\dfrac{1}{{{p^{(x)}}}}}}} \right)\], according to the property of \[{\left( {{x^a}} \right)^b}\, = \,{x^{ab}}\].
\[ \Rightarrow {\log _p}\,{\log _{p\,}}{p^{\dfrac{1}{{{p^n}}}}}\]
\[ \Rightarrow {\log _p}\dfrac{1}{{{p^n}}}\left( {{{\log }_p}p} \right)\] (according to \[{\log _x}{x^a} = a{\log _x}x\]).
\[ \Rightarrow {\log _p}\dfrac{1}{{{p^n}}}(1)\] (\[{\log _x}x = 1\]).
\[ \Rightarrow {\log _p}1 - {\log _p}{p^n}\]
\[ \Rightarrow 0 - n{\log _p}p\]
\[ \Rightarrow 0 - n\]
\[ \Rightarrow \, - n\].

Thus option A is correct, which is Independent of p.

Note: Without simplifying the ‘p’ part and using properties of the question is possible but it is way long and hard. So, this method is the best suitable method.