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Answer 1

Given Question:

Find the value of:

$\sf{\longrightarrow\:9^\dfrac{-3}{2}}$

Solution:

The base of the given question is 9. This can be further simplified as 3^2.

Substituting, we get:

$\sf{\longrightarrow\:(3^2)^\dfrac{-3}{2}}$

$\sf{\longrightarrow\:3^{2\times\dfrac{-3}{2}}$

[Identity applied = a^m x a^n = a^(mn)]

Now, cancelling both the 2s, we are left with:

$\sf{\longrightarrow\:3^{-3}}$

$\sf{\longrightarrow\:\left(\dfrac{1}{3}\right)^3}$

[Identity applied = a^(-m) = (1/a)^m]

$\sf{\longrightarrow\:\dfrac{1}{27}}$

The required answer is 1/27.

Attached below are the essential exponential laws required to solve questions containing exponents.

Knowledge Bytes:

- A number is said to be expressed in powers if it contains a base and an exponent.

- All numbers can be expressed as powers raised to one.

- Any number raised to 0 is 1.

- Always try to remove negative exponents (if any) by taking the reciprocal of the base.

Answer 2

Explanation:

Given Question:

Find the value of:

$\sf{\longrightarrow\:9^\dfrac{-3}{2}}$

Solution:

The base of the given question is 9. This can be further simplified as 3^2.

Substituting, we get:

$\sf{\longrightarrow\:(3^2)^\dfrac{-3}{2}}$

$\sf{\longrightarrow\:3^{2\times\dfrac{-3}{2}}$

[Identity applied = a^m x a^n = a^(mn)]

Now, cancelling both the 2s, we are left with:

$\sf{\longrightarrow\:3^{-3}}$

$\sf{\longrightarrow\:\left(\dfrac{1}{3}\right)^3}$

[Identity applied = a^(-m) = (1/a)^m]

$\sf{\longrightarrow\:\dfrac{1}{27}}$

The required answer is 1/27.

Attached below are the essential exponential laws required to solve questions containing exponents.

Knowledge Bytes:

- A number is said to be expressed in powers if it contains a base and an exponent.

- All numbers can be expressed as powers raised to one.

- Any number raised to 0 is 1.

- Always try to remove negative exponents (if any) by taking the reciprocal of the base.