Question

How do you simplify ${{\left[ \dfrac{1}{{{h}^{-2}}} \right]}^{-1}}.{{h}^{3}}?$

Answer 1

Hint: As we have the given equation it is in the negative form so here we have to simplify the given function by property of negative exponent due to that we can easily deal with the given function. It appears that exponential function is defined not only for natural numbers but also for rational, negative numbers. So we have to simplify it by using; ${{x}^{-n}}=\dfrac{1}{{{x}^{n}}}$

Complete step by step solution:
Given that,
${{\left[ \dfrac{1}{{{h}^{-2}}} \right]}^{-1}}.{{h}^{3}}?$
Now, we have to use here,
The exponent rules they are,
$\dfrac{1}{{{x}^{-n}}}={{x}^{n}};{{x}^{-m}}=\dfrac{1}{{{x}^{m}}}$
$\Rightarrow {{\left( {{x}^{m}} \right)}^{n}}={{x}^{mn}};{{x}^{n}}.{{x}^{m}}={{x}^{n+m}}$
$\therefore {{\left( \dfrac{1}{{{h}^{-2}}} \right)}^{-1}}.{{h}^{3}}$
$\Rightarrow $${{\left( {{h}^{2}} \right)}^{-1}}.{{h}^{3}}$
$\Rightarrow $${{h}^{-2}}.{{h}^{3}}$
$\Rightarrow $${{h}^{-2+3}}$$\left( \because {{x}^{n}}{{x}^{m}}={{x}^{n+m}} \right)$
$\Rightarrow $${{h}^{1}}$
$\therefore {{\left( \dfrac{1}{{{h}^{-2}}} \right)}^{-1}}.{{h}^{3}}=h$
Hence, we have simplified the given function.

Additional Information:
The negative exponents indicate the reciprocal of base is to be used. Let’s take the exponent as $\left( -1 \right)$ So,${{x}^{-1}}=\dfrac{1}{x}$. Now we use the power law of exponent to extend that property to any negative exponent. As an example take as $\left( {{x}^{-2}} \right)$
$\Rightarrow $${{x}^{-2}}={{\left( {{x}^{-1}} \right)}^{2}}={{\left( \dfrac{1}{x} \right)}^{2}}=\dfrac{1}{{{x}^{2}}}$
In the first way we take square to $\dfrac{1}{x}$ and then we take the inverse of ${{x}^{2}}.$By using both we get the same results. The general formula for negative exponent is ${{x}^{-a}}=\dfrac{1}{{{x}^{a}}}$
Also we take one example for better understanding $\therefore {{9}^{\dfrac{-3}{2}}}={{\left( \sqrt{9} \right)}^{-3}}={{3}^{-3}}={{27}^{-1}}=\dfrac{1}{27}$
Here the square root of the inverse of $9$ will be cubed. The inverse of $8$ cube of square root of $9.$

Note: The number less than zero is said to be the negative number. The negative number is always expressed with the negative sign. For better understanding as an example we take $\left( -4 \right)$ as we can see that it is the negative term so the given number four is less than zero and as we solve any roots of the number if the number is negative so the answer we get is also negative. As an example we take$-{{4}^{3}}=-64$ the cube root we get is negative. There are various identities which are applicable for the various equations. These are some key points you need to remember while solving these problems.