NCERT Solutions for Chapter: Exponents and Powers, Exercise 1: Exercise

Find the value of ${\left(2^4+2^3\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$.

If $abc=0$, then find the value of ${\left[{\left(x^a\right)}^b\right]}^c$.

If $x^y\times y^x=800$, then find $x-y$ $($where $x$ and $y$ are positive integers and $x>y)$.

${\left(\frac{a^{x^2}}{a^{y^2}}\right)}^{\frac{1}{x+y}}\times {\left(\frac{{\displaystyle a^{y^2}}}{{\displaystyle a^{z^2}}}\right)}^{\frac{{\displaystyle 1}}{{\displaystyle y+z}}}\times {\left(\frac{{\displaystyle a^{z^2}}}{{\displaystyle a^{x^2}}}\right)}^{\frac{{\displaystyle 1}}{{\displaystyle z+x}}}=$

If ${\left(x^2-y^2\right)}^4=256$ and ${\left(x^2+y^2\right)}^5=243$, then find $x^4-y^4$.

The following steps are involved in solving the above problem. Arrange them in sequential order.

$\left(A\right) {\left(x^2-y^2\right)}^4=256=4^4$ and ${\left(x^2+y^2\right)}^5=243=3^5$

$\left(B\right) x^4-y^4=12$

$\left(C\right) \left(x^2-y^2\right)\left(x^2+y^2\right)=4\times 3$

$\left(D\right) x^2-y^2=4 \text{and} x^2+y^2=3$

${\left(\frac{a^{x^4}}{a^{y^4}}\right)}^{\frac{1}{x^2+y^2}}\times {\left(\frac{{\displaystyle a^{y^4}}}{{\displaystyle a^{z^4}}}\right)}^{\frac{{\displaystyle 1}}{{\displaystyle y^2+z^2}}}\times {\left(\frac{{\displaystyle a^{z^4}}}{{\displaystyle a^{x^4}}}\right)}^{\frac{{\displaystyle 1}}{{\displaystyle z^2+x^2}}}=$

Arrange the following in descending order $3^{34}, 2^{51}, 7^{17}$ we get

If $p^x=r^y=m$ and $r^w=p^z=n$, then which one of the following is correct?