Question

Convert the following into power notations.
(a) -8/27
(b)49/81
(c) 8/125​

Answer 1

$\underline{\pink { Power \: notations: }}$

$\red{ a) \frac{-8}{27}}$

$= \Big( \frac{(-2)^{3}}{3^{3}}\Big)$

$= \Big( \frac{-2}{3}\Big)^{3}$

Therefore.,

$\red{ a) \frac{-8}{27}} \green { = \Big( \frac{-2}{3}\Big)^{3}}$

$\red{ b) \frac{49}{81}}$

$= \Big( \frac{(7)^{2}}{9^{2}}\Big)$

$= \frac{7^{2}}{3^{4}}$

Therefore.,

$\red{ a) \frac{49}{81}} \green { = \frac{7^{2}}{3^{4}}}$

$\red{ c) \frac{8}{125}}$

$= \Big( \frac{(2)^{3}}{5^{3}}\Big)$

$= \Big( \frac{2}{5}\Big)^{3}$

Therefore.,

$\red{ a) \frac{8}{125}} \green { = \Big( \frac{2}{5}\Big)^{3}}$

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Answer 2

To find: The power notations of the following:

(a) $-\dfrac{8}{27}$

(b) $\dfrac{49}{81}$

(c) $\dfrac{8}{125}$

Identity : $\dfrac{a^n}{b^n}=(\dfrac{a}{b})^n$

Solution: (a) Since $8=2\times2\times2=2^3$

$27=3\times3\times3=3^3\\\\\Rightarrow\ -\dfrac{8}{27}=-\dfrac{2^3}{3^3}=-(\dfrac{2}{3})^3$

(b)  Since $49=7\times7=7^2,\ \ 81= 9\times9=9^2$

$\Rightarrow\ \dfrac{49}{81}=\dfrac{7^2}{9^2}=(\dfrac{7}{9})^2$

(c) Since $8=2\times2\times2=2^3$

$125=5\times5\times5=5^3\\\\\Rightarrow\ \dfrac{8}{125}=\dfrac{2^3}{5^3}=(\dfrac{2}{5})^3$

Hence, (a)$-\dfrac{8}{27}=-(\dfrac{2}{3})^3$  (b)$\dfrac{49}{81}=(\dfrac{7}{9})^2$   (c) $\dfrac{8}{125}=(\dfrac{2}{5})^3$