Question

#BrainTeaser02

Find the possible values of n.
$(n^2)^n - 2n^n+ 1 = 0 ~~~~~~~(n \neq 0,1)$

Answer 1

$\sf{(n^2)^n - 2n^n+ 1 = 0 ~~~~~~~(n \neq 0,1)}$

$\sf{\fbox{let \: n = 2}}$

$\sf{(2^2)^2 - 2(2)^2+ 1 = 0} \\ \sf{ {2}^{4} - 2(4) + 1 = 0} \\ \sf{16 - 8 + 1 = 0} \\ \sf \purple{9 \neq0} \\ \sf \red{hence, \: n \neq2}$

$\tt{after \: taking \: a \: few \: values,}$

$\sf \pink{let \: n \neq 0,1,2, - 2, - 3, - 4, 4}$

Answer 2

Required Answer

$\textsf{We have \sf{4\left(n^{2}\right)^{n}-2 n^{n}+1=0} and n \neq 0, 1}$

$\to\textsf{We can write \sf{(n^2)^n} as \sf{(n^n)^2} }$

$\to\textsf{So, \sf{\left(n^{n}\right)^{2}-2 n^{n}+1=0} }$

$\bigstar$ $\bf{Let\:n^n=x}$

$\implies\sf{x^{2}-2 x+1=0}$

$\implies\sf{(x-1)^2=0}$

$\implies\sf{x=1}$

• $\sf{So,\:n^n=1}$

$\textbf{Taking log on both sides}$

$\implies\sf{n \log (n)=\log 1}$

$\implies\sf{\log (n)=\frac{0}{n}}$

$\implies\sf{\log (n)\neq 0\quad (since, n\neq 0 )}$

$\implies\sf{\log (n)=\log (1)}$

$\implies\sf{n=1}$

$\textsf{But from the question n \neq 1}$

$\textbf{We can not find value of "n" other than 1.}$