Now,

$ \Rightarrow 6561 = {3^8}$

${\left( {6561} \right)^{0.25}} = {\left( {6561} \right)^{1/4}}$

$ = {\left( {{3^8}} \right)^{1/4}}$

$ = {\left( 3 \right)^{8/4}}$

$ = {3^2} = 9$

Given,

We have,

Given that:

Given $x=y^{\dfrac{1}{a}}.....(1)$

The value of

  $ \dfrac{{{100}^{98}}+{{100}^{100}}}{{{100}^{98}}}+1 $

 $ =\dfrac{{{100}^{98}}}{{{100}^{98}}}+\dfrac{{{100}^{100}}}{{{100}^{98}}}+1 $

 $ =1+{{100}^{100-98}}+1 $

 $ ={{100}^{2}}+2 $

 $ =10000+2 $

 $ =10002 $

Hence, this is the answer. 

We have,
$(3r^2)\times (9r^2)^{3/2}\div(27r^{-3})^{1/3}\$

Now,

$\displaystyle \left ( 512 \right )^{\frac{-2}{9}}=\left ( 2^{9} \right )^{\frac{-2}{9}}=2^{-2}=\frac{1}{4}$

$={8}^{3+2-5}$

$1) 3^{2^{2^{2^{2}}}}=3^{2^{2^{4}}}=3^{16}$

$4^{2x}=\frac {1}{32}\Rightarrow (2^2)^{2x}=\frac {1}{2^5}$
$\Rightarrow 2^{4x}=2^{-5}$
$\Rightarrow 4x=-5$
$x=\frac {-5}{4}$

Given that:

The factors $2$ and $5$ in the given number are the factors of $10$.
We can write $N$ as-
$ N=2^8\times2^4\times5^8 $
$=2^4(2^8\times5^8)$
$=2^4(10)^8=16\times10^8$
Therefore, the number of digits in $N$ is $10$.

$x^1/2 / x^3/2$

$1 \times \dfrac{9}{4} - 1 \times \dfrac{3}{4}$

correct option is A..

$=\dfrac {x^{5/6}}{x^{11/6}}$

If the division of two bases is powered by the same exponent,

If the product of the bases is powered  by the same exponent, 

$\left (\dfrac {2^{3}}{3^{3}} \right )^{2} = \dfrac {2^{6}}{3^{6}} = \dfrac {64}{729}$

$\displaystyle 3^{2}\times a^{3}\times b^{3}=\left ( 3ab \right )^{3}$.

$\displaystyle \left [ \left ( -1 \right )^{2}\times \left ( -1 \right )^{3}\times \left ( -1 \right )^{4} \right ]^{6}$

$\displaystyle \left ( -4 \right )^{3}\times \left ( -3 \right )^{3}=\left ( -4\times -3 \right )^{3}$

$\displaystyle a^{x}\times b^{x}=\left ( ab \right )^{n}$

$\displaystyle 5^{2}\times 3^{2}=\left ( 15 \right )^{2}$

=$\displaystyle \left [ \left ( 4 \right )^{\tfrac{1}{4}}\times \left ( 2 \right )^{\tfrac{1}{2}} \times \left ( 5 \right )^{\tfrac{1}{2}}\right ]^{0}$

$\displaystyle \left [ \left ( \frac{-2}{5} \right )^{3} \right ]^{2}=\left ( \frac{-2}{5} \right )^{6}=\frac{\left ( -2 \right )^{6}}{5^{6}}=\frac{64}{15625}$

$(-a)^9 \times (-b)^9 = [ -a \times -b]^9$

$\displaystyle\left [ \frac{-3}{7} ^{-1}\right ]^{2}=\left [ \left ( \frac{-7}{3} \right )^{1} \right ]^{2}=\left ( \frac{-7}{3} \right )^{2}=\frac{49}{9}$

$\displaystyle\left ( \frac{2}{3} \right )^{-5}=\left ( \frac{3}{2} \right )^{5}=\frac{3^{5}}{2^{5}}=\frac{243}{32}$

$\displaystyle\left ( 16\div 15 \right )^{3}=16^{3}\div 15^{3}$ 

$\displaystyle \frac{a^{m}}{a^{m}}=1$
$\displaystyle\therefore  \frac{a^{m}}{a^{m}}\neq a^{m}$

Let $x=296q+75$
$=(37\times 8q+37\times 2)+1$
$=37(8q+2)+1$
Thus, when the number is divided by $37$, the remainder is $1$.

Let $n=4q+3$. Then $2n=8q+6=4(2q+1)+2$
Thus, when $2n$ is divided by $4$, the remainder is $2$.

Given Exp. = $\dfrac{1}{\begin{pmatrix}1+\dfrac{x^b}{x^a}+\dfrac{x^c}{x^a}\end{pmatrix}} + \dfrac{1}{\begin{pmatrix}1+\dfrac{x^a}{x^b}+\dfrac{x^c}{x^b}\end{pmatrix}} + \dfrac{1}{\begin{pmatrix}1+\dfrac{x^b}{x^c}+\dfrac{x^a}{x^c}\end{pmatrix}}$
$= \dfrac{x^a}{(x^a+x^b+x^c)}+\dfrac{x^b}{(x^a+x^b+x^c)} + \dfrac{x^c}{(x^a+x^b+x^c)}$
$= \dfrac{x^a+x^b+x^c}{(x^a+x^b+x^c)}$
$= 1.$

$(10)^{150}\div (10)^{146} = \dfrac{10^{150}}{10^{146}}$
$= 10^{150-146}$
$= 10^4$
$= 10000.$

$(256)^{0.16}\times (256)^{0.09} = (256)^{(0.16+0.09)}$
$= (256)^{0.25}$
$= (256)^{(25/100)}$
$= (256)^{(1/4)}$
$= (4^4)^{(1/4)}$
$= 4^{4(1/4)}$
$= 4^1$
$= 4$

we know,

As we know that $a^{m}\times a^{n}=a^{(m+n)}$

As we can see that the power is even and we know that when $(-1)$ is raised to some odd power then its value remains same i.e. $-1$ but when the power is even its sign is changed i.e. $1$ . therefore, $(-1)^{50}=1$

$Now\quad \left[ \dfrac { 100 }{ 101 }  \right] ^{ 3 }\quad \ \quad \quad =\quad \dfrac { { 10 }0^{ 3 } }{ 101^{ 3 } } \quad \left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) \ $

$Now\quad \dfrac { { 10 }^{ 2 } }{ 11^{ 2 } } \ =\quad \left( \dfrac { 10 }{ 11 }  \right) ^{ 2 }\left( \because \left( \dfrac { { a }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a }{ b }  \right) ^{ m } \right) $

$Now\quad \left( \dfrac { { 5 }^{ 5 }\times { 6 }^{ 5 } }{ { 3 }^{ 5 } }  \right) \quad \ \quad \quad =\quad \left( \dfrac { 5\times 6 }{ 3 }  \right) ^{ 5 }\quad \left( \because \left( \dfrac { { a }^{ m }\times { c }^{ m } }{ { b }^{ m } }  \right) =\left( \dfrac { a\times c }{ b }  \right) ^{ m } \right) $

we know,

we know,

$(\sqrt2)^x+(\sqrt3)^x=(\sqrt{13})^{\frac{x}{2}}$

$a^2bc^3=5^3$   ......(1)

$225=3^2.5^2=5^{x-3}.3^{2x-a}$

To rationalize $(a^2b^3c^4)^{\frac{1}{5}}$, fifth root must be removed, 

We have, $\Bigr(\dfrac{5^a}{5^b}\Bigl)^{a+b}\cdot\Bigl(\dfrac{5^b}{5^c} \Bigr)^{b+c}\cdot \Bigl(\dfrac{5^c}{5^a} \Bigr)^{c+a}$

${ 10 }^{ -49 }-2\cdot { 10 }^{ -50 }={ 10 }^{ -50 }(10-2)=8\cdot { 10 }^{ -50 }\ \therefore { 10 }^{ -49 }\hspace{1mm} exceeds\hspace{1mm} 2\cdot { 10 }^{ -50 }\hspace{1mm} by\hspace{1mm} 8\cdot { 10 }^{ -50 }$

$\therefore a.b= \dfrac { \log _{ 10 }{ 3 }  }{ \log _{ 10 }{ 2 }  }   \times \dfrac { \log _{ 10 }{ { 2 } }  }{ \log _{ 10 }{ { 3 } }  }$

$2^{n-m}=16\2^{n-m}=2^4\n-m=4\cdots(1)\3^{n+m}=729\3^{n+m}=3^6\n+m=6\cdots(2)\(1)+(2)\2n=10\\boxed{n=5}\\boxed{m=1}$
$\boxed{nm=5}...Answer$

$(\frac{\sqrt[3]{180}\times \sqrt[6]{4}}{\sqrt[4]{81}})\\ ((\frac{27\times4)^{(\frac{1}{4})}\times 2^{(\frac{2}{6})}}{(3^4)^{(\frac{1}{4})}}))\\ =(\frac{3\times 2^{(\frac{2}{3}) }\times 2^{(\frac{1}{3})}}{3})\\= 2^{{(\frac{2}{3})}+{(\frac{1}{3})}}\\=2$

Given, 

$(32)(32)^{1/6}(32)^{1/36}......$ to $\infty = (32)^{1+\dfrac{1}{6}.....\infty}$

$\begin{array}{l}\left[ {{{\left( {{{\left( {625} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right] = \left[ {{{\left( {{{\left( {{{25}^2}} \right)}^{\frac{{ - 1}}{2}}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = \left[ {{{\left( {{{25}^{ - 1}}} \right)}^{\frac{{ - 1}}{4}}}} \right]\ = {25^{\frac{1}{4}}}\ = {5^{2 \times \frac{1}{4}}}\ = {5^{\frac{1}{2}}}\ = \sqrt 5 \end{array}$

Substitute the value of $n=3$ in $b$
 $\Rightarrow  3n^2+3=3\times 3^2+3=27+3= 30 $
Hence, option 'A' is correct.

$\displaystyle (64) ^{-\tfrac{1}{2}}-(-32)^{-\tfrac{4}{5}}$
$=\left({2^6}\right)^{\frac{-1}{2}}-\left({(-2)^5}\right)^{\frac{-4}{5}}$
$=(2)^{ -\tfrac { 6 }{ 2 }  }-(-2)^{ -\tfrac { 4\times 5 }{ 5 }  }$
$=(2)^{ -3 }-(-2)^{ -4 }$
$\dfrac { 1 }{ 8 } -\dfrac { 1 }{ 16 } =\dfrac { 1 }{ 16 } $
Answer $C$ option, $ \cfrac { 1 }{ 16 } $

$a^x=\sqrt{b}=b^{1/2}$
$a=b^{1/2x}$
$b^y=\sqrt[3]{c}$
$b^y=c^{1/3}$
$b=c^{1/3y}$
$c^z=a^{1/2}$
$c=a^{1/2z}=b^{1/4xz}$
$c^1=c^{1/12xyz}$
$\displaystyle 1= \frac {1}{12xyz}$
$\displaystyle xyz = \frac {1}{12}$

 $\displaystyle \frac{2^{x-3}}{8^{-x}} = \frac{32}{4^{(1/2)x}}$
$\frac { 2^{ x-3 } }{ 2^{ -3x } } =\frac { { 2 }^{ 5 } }{ 2^{ (2/2)x } } $\ 
${ 2 }^{ x-3+3x }={ 2 }^{ 5-x }$
$4x-3=5-x$
$5x=8$
$x=1\frac { 3 }{ 5 } $
Answer (D)  $1\frac { 3 }{ 5 } $

$2^a\,>\,4^c\;\;\;\;\;\;3^b\,>\,9^a$
$2^a\,>\,2^{2c}\;\;\;\;\;3^b\,>\,3^{2a}$
$a\,>\,2c\;\;\;\;\;\;\;b\,>\,2a$
$\therefore\;a\,>\,c-(i)\;\;\therefore\;b\,>\,a-(ii)$
From (1) & (2), we have
$c\,<\,a\,<\,b$

$[(-2)^{3} \times (-2)^{-4}]^{2} = [(-2)^{3-4}]^{2}$

$\displaystyle \left ( \frac{11^{2}}{13^{2}} \right )^{-6}=\left ( \frac{11}{13} \right )^{-12}=\left ( \frac{13}{11} \right )^{12}$
Also,$\displaystyle \left ( \frac{13}{11} \right )^{12}=\left ( \frac{13}{11} \right )^{m}$
$\displaystyle \therefore m=12$

$[(-3)^{-4} \div (-3)^{-5}]^{3} = [(-3)^{-4+5}]^{3}$

$(6^{4} \times 7^{2})^{\tfrac {1}{2}} = 6^{\tfrac {4}{2}} \times 7^{\tfrac {2}{2}}$
$= 6^{2} \times 7$
$= 36\times 7$
$= 252$

$\log _{10} x = a$

$10^a = x$

$\therefore 10^{a – 1} = \dfrac{10^a}{10^ 1} = \dfrac{x}{10}$

$\log _{10} y = b$

$10^b = y$

 $ \therefore 10^{2b} = (10^b)^2 = y^2$

We know that for some positive integers $m$ and $n$, 

we know that,

As we know that $a^{-b}$ is equal to $1/a^{b}$. Also, $p^{0}=1$ 

$(x^{q-r})^{\cfrac{1}{qr}}\times (x^{r-p})^{\cfrac{1}{rp}}\times (x^{p-q})^{\cfrac{1}{pq}} $

We can write the given equation as, 

$100th $ root of $10^{(10^{10})}=(10^{(10^{10})})^{\frac{1}{100}}$

$a^0=1\,\,(\text{Where} \,a\ne 0)$  

Given, $16^{(x^2 + 3x -1)} = 8^{(x^2 + 3x + 2)}$