Question

Answer fast I will mark as brainliest. No nonsense answers. Give step by step explanation. If u will not, I will report

Answer 1

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

Given:

$\sf = \bigg(-\dfrac{2}{3} \bigg)^{10} \times \bigg(-\dfrac{2}{3} \bigg)^{ - 15} \div \bigg(-\dfrac{2}{3} \bigg)^{25}$

As we know that:

$\sf \implies {x}^{a} \times {x}^{b} = {x}^{a + b}$

$\sf \implies {x}^{a} \div {x}^{b} = {x}^{a - b}$

We get:

$\sf = \bigg(-\dfrac{2}{3} \bigg)^{10 - 15} \div \bigg(-\dfrac{2}{3} \bigg)^{25}$

$\sf = \bigg(-\dfrac{2}{3} \bigg)^{ - 5} \div \bigg(-\dfrac{2}{3} \bigg)^{25}$

$\sf = \bigg(-\dfrac{2}{3} \bigg)^{ - 5 - 25}$

$\sf = \bigg(-\dfrac{2}{3} \bigg)^{ -30}$

$\sf = \bigg(\dfrac{2}{3} \bigg)^{ -30} \: \: \: (Answer)$

$\texttt{\textsf{\large{\underline{Know More}:}}}$

If a, b are positive real numbers and m, n are rational numbers, then the following results hold -

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

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

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

$\sf4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\sf5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\sf6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\sf7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\sf8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$