Question

(25) power -3/2 -5 power-3.



Can you solve this one​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given expression.

$\rm = {7}^{0} \times {(25)}^{^{ -3}/_{2}} - {5}^{ - 3}$

We know that:

$\rm: \longmapsto {x}^{0} = 1,x \neq0$

Therefore, we get:

$\rm = 1 \times {(25)}^{^{ -3}/_{2}} - {5}^{ - 3}$

$\rm = {(25)}^{^{ -3}/_{2}} - {5}^{ - 3}$

$\rm = {( {5}^{2} )}^{^{ -3}/_{2}} - {5}^{ - 3}$

$\rm = {5}^{2 \times ^{ -3}/_{2}} - {5}^{ - 3}$

$\rm = {5}^{ - 3} - {5}^{ - 3}$

$\rm = 0$

★ Hence, the required answer is 0.

$\textsf{\large{\underline{Learn More}:}}$

Laws Of Exponents: If a, b are positive real numbers and m, n are rational numbers, then the following results hold.

$\rm 1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm 2. \: \: ({a}^{m})^{n} = {a}^{mn}$

$\rm 3. \: \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\rm4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\rm5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\rm6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\rm7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\rm8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$