Name: Zero & Negative Integral Indices
Uploaded: 2019-07-19
Duration: 3 min 8 s
Description: When it comes to zero and negative integral indices, there’s nothing like taking apples in baskets to figure out how it’s done. Let's watch and learn.

Question

Evaluate:

${(\frac{25}{4})}^{\frac{- 1}{2}} - {(\frac{1}{27})}^{\frac{1}{3}} + 7^{0}$

Accepted Answer

Given expression is ${(\frac{25}{4})}^{\frac{- 1}{2}} - {(\frac{1}{27})}^{\frac{1}{3}} + 7^{0}$ .

We need to evaluate the given expression.

Thus, ${(\frac{25}{4})}^{\frac{- 1}{2}} - {(\frac{1}{27})}^{\frac{1}{3}} + 7^{0}$

$\Rightarrow {(\frac{5^{2}}{2^{2}})}^{\frac{- 1}{2}} - {(\frac{1}{3^{3}})}^{\frac{1}{3}} + 1$ $[∴ a^{0} = 1, when a \neq 0]$

$\Rightarrow {(\frac{5}{2})}^{2 \times \frac{- 1}{2}} - {(3^{- 3})}^{\frac{1}{3}} + 1$ $[∴ a^{- n} = \frac{1}{a^{n}}]$

$\Rightarrow {(\frac{5}{2})}^{- 1} - {(3)}^{- 3 \times \frac{1}{3}} + 1$ $[∴ {(a^{n})}^{m} = a^{n m}]$

$\Rightarrow (\frac{2}{5}) - {(3)}^{- 1} + 1$ $[∴ {(\frac{a}{b})}^{- n} = {(\frac{b}{a})}^{n}]$

$\Rightarrow \frac{2}{5} - \frac{1}{3} + 1$ $[∴ a^{- n} = \frac{1}{a^{n}}]$

$\Rightarrow \frac{2}{5} + 1 - \frac{1}{3}$

$\Rightarrow \frac{7}{5} - \frac{1}{3} = \frac{21 - 5}{15} = \frac{16}{15} = 1 \frac{1}{15}$

Therefore, the value of ${(\frac{25}{4})}^{\frac{- 1}{2}} - {(\frac{1}{27})}^{\frac{1}{3}} + 7^{0}$ is $\frac{16}{15}$ or $1 \frac{1}{15}$ .

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