Question

simplify and give reasons.(4/7)^-5×(7/4)^-7.
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Answer 1

Question:

$( \frac{4}{7} ) {}^{ - 5} \times ( \frac{7}{4} ) {}^{ - 7}$

Solution:

$( \frac{4}{7} ) {}^{ - 5} \times ( \frac{7}{4} ) {}^{ - 7}$

Property to be used :

$a {}^{ - x} = ( \frac{1}{a} ) {}^{x}$

So,

$( \frac{4}{7} ) {}^{ - 5} \times ( \frac{7}{4} \: ^{7} )$

Property :

$a {}^{x} \times a {}^{y} = a {}^{x + y}$

So,

using the property

$( \frac{4}{7} ) {}^{ - 5 + 7}$

$= ( \frac{4}{7} ) {}^{2}$

Hence,

$( \frac{4}{7} ) {}^{ - 5} \times ( \frac{7}{4} ) {}^{ - 7}$

$= ( \frac{4}{7} ) {}^{2}$

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Extra :

Two basic properties used to solved this question

$1) \: \: a {}^{ - x} = ( \frac{1}{a} ) {}^{x}$

$2) \: a {}^{x} \times a {}^{y} = a {}^{x + y}$

Answer 2

Question

$\sf (\frac{4}{7})^{-5}\times (\frac{7}{4})^{-7}$

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Answer

$\sf (\frac{4}{7})^{-5}\times (\frac{7}{4})^{-7}$

Step 1: Apply this law of exponent ⇒ $\sf a^{-m}=\frac{1}{a^{m}}$

$\sf (\frac{4}{7})^{-5}\times (\frac{7}{4})^{-7}$

$\sf (\frac{4}{7})^{-5}\times (\frac{4}{7})^{7}$

Step 2: Apply this law of exponent ⇒ $\sf a^{m} \times a^{n} = a^{m+n}$

$\sf (\frac{4}{7})^{-5}\times (\frac{4}{7})^{7}$

$\sf (\frac{4}{7})^{-5+7}$

$\sf (\frac{4}{7})^{2}$

Step 3: Find the actual value.

$\sf (\frac{4}{7})^{2}$

$\sf \frac{4\times 4 }{7\times 7 }$

$\sf \frac{16}{49}$

$\bf \therefore \ (\frac{4}{7})^{-5}\times (\frac{7}{4})^{-7} = \frac{16}{49}$

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Additional Information

What are exponents?

Something that shows repeated multiplication is known as an exponent. Another name for exponents is powers.

For Example ⇒ $\sf 3^2 = 3 \times 3$

What are the parts of an exponential number?

An exponential number consists of two parts. The base and the exponent part.

It is like this ⇒ $\sf Base^{Exponent}$

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