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

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.