Question

How do you solve \[{x^{ - \dfrac{2}{3}}} = 9\]?

Answer 1

Hint: Here in this question, we have to find the value of x. This can be solved by two methods. Firstly, by raising the power on both sides by simplification using the law of indices we get the required solution otherwise it can also be solved by using the radicand simplification to get the required x value.

Complete step-by-step answer:
Consider the given expression
\[{x^{ - \dfrac{2}{3}}} = 9\]---------(1)
Here, we have to find the value of x.
Method:1
To solve x, raise the power or multiplying power \[ - \dfrac{3}{2}\] on both sides in equation (1), then
\[ \Rightarrow {x^{\left( { - \dfrac{2}{3}} \right)\left( { - \dfrac{3}{2}} \right)}} = {9^{ - \dfrac{3}{2}}}\]
On by simplification, we get
 \[ \Rightarrow x = {9^{ - \dfrac{3}{2}}}\]
By the law of indices \[{x^{\dfrac{n}{m}}} = \sqrt[m]{{{x^n}}}\], RHS can be written as
\[ \Rightarrow x = \sqrt {{9^{ - 3}}} \]
As we know 9 is a square number of 3, then
\[ \Rightarrow x = \sqrt {{{\left( {{3^2}} \right)}^{ - 3}}} \]
Or
\[ \Rightarrow x = \sqrt {{{\left( {{3^{ - 3}}} \right)}^2}} \]
On by simplification, we get
\[ \Rightarrow x = {3^{ - 3}}\]
Again, by law of indices \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\], then
\[ \Rightarrow x = \dfrac{1}{{{3^3}}}\]
The cube value of 3 is 27, i.e.,
\[ \Rightarrow x = \dfrac{1}{{27}}\]
Method:2
By using the law of indices \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\], LHS of equation (1) can be written as
\[ \Rightarrow \dfrac{1}{{{x^{\dfrac{2}{3}}}}} = 9\]
Again, by using the law of indices \[{x^{\dfrac{n}{m}}} = \sqrt[m]{{{x^n}}}\] denominator of RHS can be written as
\[ \Rightarrow \dfrac{1}{{{{\left( {\sqrt[3]{x}} \right)}^2}}} = 9\]
on cross multiplication, we get
\[ \Rightarrow \dfrac{1}{9} = {\left( {\sqrt[3]{x}} \right)^2}\]
On rearranging
\[ \Rightarrow {\left( {\sqrt[3]{x}} \right)^2} = \dfrac{1}{9}\]
Taking square root on both side, then
\[ \Rightarrow \sqrt[3]{x} = \pm \sqrt {\dfrac{1}{9}} \]
As we know 9 is the square number of 3, then
\[ \Rightarrow \sqrt[3]{x} = \pm \dfrac{1}{{\sqrt {{3^2}} }}\]
On simplification, we get
\[ \Rightarrow \sqrt[3]{x} = \pm \dfrac{1}{3}\]
Taking cube on both side, then
\[ \Rightarrow x = \pm {\left( {\dfrac{1}{3}} \right)^3}\]
As we know \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\], then RHS can be written as
\[ \Rightarrow x = \pm \dfrac{1}{{{3^3}}}\]
Cube value of 3 is 27
\[ \Rightarrow x = \pm \dfrac{1}{{27}}\]
Hence, the value of x is \[x = \dfrac{1}{{27}}\].
So, the correct answer is “\[x = \dfrac{1}{{27}}\]”.

Note: The question is based on the exponential form. To solve this question we have used simple arithmetic operations. Since the power has 2 and 3 we must know about the square and cube root of the number. The law of indices is used whenever it is necessary and solved the equation.