Question

How do you simplify \[{9^{ - \dfrac{3}{2}}}\] ?

Answer 1

Hint: For solving this question, we have to learn about the power of coefficients. Like ${2^2} = 4$, ${2^3} = 8$ and so on. We also need to know about the square root and cube root of a number like $\sqrt 9 = 3$ , $\sqrt[3]{{27}} = 3$ . Remember, ${27^{\dfrac{1}{3}}}$ , $\sqrt[3]{{27}}$ and ${(27)^{\dfrac{1}{3}}}$ all are same.

Complete step by step solution:
The given number to simplify is \[{9^{ - \dfrac{3}{2}}}\]
Now, we will break ${9^{ - \dfrac{3}{2}}}$ down into its component parts$ - \dfrac{3}{2}$ is the same as $3 \times - \dfrac{1}{2}$ which is the same as $\dfrac{1}{2} \times - 3$
So ${9^{ - \dfrac{3}{2}}}$ is the same as $\dfrac{1}{{{9^{\dfrac{3}{2}}}}}$
Similarly,$\dfrac{1}{{{9^{\dfrac{3}{2}}}}}$ can also be written as $\dfrac{1}{{{{({9^{\dfrac{1}{2}}})}^3}}}$
Now,$\dfrac{1}{2} \times 2$ will cancel out in the power of 9
Now, we get that
$ = \dfrac{1}{{{3^3}}}$
We know that 3 to the power 3 is 27
$ = \dfrac{1}{{27}}$.

So, when we simplify \[{9^{ - \dfrac{3}{2}}}\]we get $\dfrac{1}{{27}}$.

Additional Information:
In taking the square root of 9 we only took the primary root; a secondary solution exists $\dfrac{1}{{{{( - 3)}^3}}} = - \dfrac{1}{{27}}$ . The square root of any real number has both the plus and minus sign. For example, \[\sqrt[2]{4} = \pm 2\]. This is because the square of -2 or +2 would be the same, that is 4.
We need to be careful that squaring a number means you need to multiply the number two times.
For example:
\[ \Rightarrow {2^2} = 2 \times 2 = 4\]
Similarly, cubing a number we need to multiply the number three times.
For example:
$ \Rightarrow {2^3} = 2 \times 2 \times 2 = 8$
We also need to be careful that squaring and the square root is just the opposite of each other that means square root means we can multiply with itself to get the number
For example:
$ \Rightarrow \sqrt 4 = 2$ because \[{2^2} = 2 \times 2 = 4\]
Similarly, cube root means we can multiply with itself to get the number
For example:
$ \Rightarrow \sqrt[3]{8} = 2$ because ${2^3} = 2 \times 2 \times 2 = 8$.

Note: There is an alternative method for the problem \[{9^{ - \dfrac{3}{2}}}\]
Firstly, change the 9 to ${3^2}$, so we get,
$ = {({3^2})^{ - \dfrac{3}{2}}}$
Now cancel out the 2’s in the power
$ = {(3)^{ - 3}}$
We know that ${(3)^{ - 3}}$ is the same as $\dfrac{1}{{{3^3}}}$ , we get,
$ = \dfrac{1}{{27}}$
So, when we simplify \[{9^{ - \dfrac{3}{2}}}\]we get $\dfrac{1}{{27}}$.