Introduction to indices - class-VI

Numerator $=2^{m+3}\cdot3^{2m-n}\cdot5^{m+n+3}\cdot2^{n+1}\cdot3^{n+1}$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (i)
Denominator $=2^{m+1}\cdot3^{m+1}\cdot2^{n+3}\cdot5^{n+3}\cdot3^m\cdot5^m$
$=2^{m+n+4}\cdot3^{2m+1}\cdot5^{m+n+3}$ (ii)
Given expression $=1$

Unit digit in ${6374}^{1993}=$ Unit digit in ${(4)}^{1793}$
=Unit digit in $[{({4}^{2})}^{896}\times 4]$
=Unit digit in $(6\times 4)=4$
Unit digit in ${(625)}^{317}=$ Unit digit in ${(5)}^{317}=5$
Unit digit in ${(341)}^{491}=$ Unit digit in ${(1)}^{491}=1$
Required digit$=$ Unit digit in $(4\times 5\times1)=0$

To find the maximum of x^(1/x), consider the function f(x) = x^(1/x). Its derivative is f'(x) = x^(1/x) * (1 - ln(x)) / x^2. The maximum occurs at x = e (approx 2.718). Comparing values near e, 3^(1/3) is larger than 2^(1/2) and 4^(1/4).

Thus, the number ends with $5$ for $n\in N$ and $n$ is odd or even.

$\displaystyle\frac{\displaystyle1+\frac{1}{5}}{\displaystyle1-\frac{1}{5}}-\left[\left(\frac{2}{3}\right)^3\right]^{\displaystyle\frac{1}{3}}-\left[\left(\frac{6}{5}\right)^2\right]^{\displaystyle-\frac{1}{2}}=\frac{\displaystyle\frac{6}{5}}{\displaystyle\frac{4}{5}}-\left(\frac{2}{3}\right)^1-\left(\frac{6}{5}\right)^{-1}=\frac{6}{4}-\frac{2}{3}-\frac{5}{6}=\frac{18-8-10}{12}=0$

${(3.92^2+3(2.1^4)}^\dfrac{1}{6}$

$\displaystyle { 2 }^{ n }-{ 2 }^{ n-1 }=4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }\left( 2-1 \right) =4$
$\displaystyle \therefore \quad { 2 }^{ n-1 }={ 2 }^{ 2 }$
$\displaystyle \therefore \quad n-1=2$
$\displaystyle \therefore \quad n=3$
$\displaystyle \therefore \quad { n }^{ n }={ 3 }^{ 3 }=27$

$\displaystyle 2^{100}\div 2^1=2^{100-1}=2^{99}$

$\displaystyle { \left( { x }^{ 6 }.{ y }^{ { -5 }/{ 4 } } \right)  }^{ { -4 }/{ 3 } }$
$\displaystyle ={ x }^{ -6\times { 4 }/{ 3 } }{ y }^{ \dfrac { -5 }{ 4 } \times -\dfrac { 4 }{ 3 }  }$
$\displaystyle ={ x }^{ -8 }{ y }^{ { 5 }/{ 3 } }$