Question

How do I do this? :////​

Answer 1

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

Let the number be multiplied be x.

Therefore, according to the given conditions:

$\rm:\longmapsto {3}^{ - 3} \times x = 4$

We know that:

$\rm:\longmapsto {x}^{ - a} = \dfrac{1}{ {x}^{a} }$

Therefore:

$\rm:\longmapsto \dfrac{1}{ {3}^{3} } \times x = 4$

$\rm:\longmapsto \dfrac{x}{27 } = 4$

Multiplying both sides by 27, we get:

$\rm:\longmapsto27 \times \dfrac{x}{27 } = 4 \times 27$

$\rm:\longmapsto x = 4 \times 27$

$\rm:\longmapsto x = 108$

★ So, 108 must be multiplied to get 4.

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

Put x = 108, we get:

$\rm = {3}^{ - 3} \times 108$

$\rm = \dfrac{1}{27} \times 108$

$\rm = \dfrac{1}{27} \times 27 \times 4$

$\rm =4$

Hence, our answer is correct (Verified)

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

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

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

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

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

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

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

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

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

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

Answer 2

Step-by-step explanation:

let the number be x,

$\sf (3^{-3})(x)=4$

$\sf \bigg( \dfrac{1}{3^3}\bigg)(x)=4$

$\sf \bigg( \dfrac{x}{27}\bigg)=4$

$\sf x=4×27$

$\sf x=108$