How do you simplify \[{\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}}\]?
Answer
599.1k+ views
Hint: We can see this problem is from indices and powers. We can solve this problem using the rule: law of brackets \[ {\left( {\dfrac{a}{b}} \right)^n} = \left( {\dfrac{{{a^n}}}{{{b^n}}}} \right)\]. We find the factors of 4 and 9 and then we express it in power of the number, this is because we have a fraction of a power in the given problem.
Complete step by step solution:
Given, \[{\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}}\].
We know that 4 can be factored as \[ \Rightarrow 4 = 2 \times 2\]
It can be written as \[ \Rightarrow 4 = {2^2}\]
Now we can factor 9 as \[ \Rightarrow 9 = 3 \times 3\]
It can be written as \[ \Rightarrow 9 = {3^2}\]
Now the given problem as
\[\Rightarrow {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}} = {\left( {\dfrac{{{2^2}}}{{{3^2}}}} \right)^{\dfrac{3}{2}}}\]
Now applying law of brackets we have,
\[\Rightarrow \dfrac{{{{\left( {{2^2}} \right)}^{\dfrac{3}{2}}}}}{{{{\left( {{3^2}} \right)}^{\dfrac{3}{2}}}}}\]
\[\Rightarrow \dfrac{{{2^{2 \times \dfrac{3}{2}}}}}{{{3^{2 \times \dfrac{3}{2}}}}}\]
\[\Rightarrow \dfrac{{{2^3}}}{{{3^3}}}\]
\[\Rightarrow \dfrac{8}{{27}}\].
Thus we have \[ {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}} = \dfrac{8}{{27}}\]
Additional information:
We know the laws of indices.
Rule 1: if a constant or variable has index as ‘0’, then the result will be equal to one, regardless of any base value. That is \[{a^0} = 1\] and ‘a’ is not equal to zero.
Rule 2: If the index is a negative value, then it can be shown as the reciprocal of the positive index raised to the same variable. That is \[{a^{ - 1}} = \dfrac{1}{a}\].
Rule 3: If we have two variables with the same base and different exponent, we add the exponent that is \[{x^a} \times {x^b} = {x^{a + b}}\].
Rule 4: If we have division of two number with same base, that is \[\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}\].
Rule 5: We also know the bracket law that is \[ {\left( {\dfrac{a}{b}} \right)^n} = \left( {\dfrac{{{a^n}}}{{{b^n}}}} \right)\]
These are some basic indices laws.
Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be molded according to our convenience while solving the problem. Also note that cube-root ( to the power \[\dfrac{1}{3}\]), square-root ( to the power \[\dfrac{1}{2}\]) are fractions with 1 as numerator and respective root in denominator but the power in our problem is first cube root then square. For example suppose if we have \[{\left( {25} \right)^{ - \left( {\dfrac{1}{2}} \right)}}\]. We can write it as \[\sqrt {25} \] which gives us the answer 5.
Complete step by step solution:
Given, \[{\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}}\].
We know that 4 can be factored as \[ \Rightarrow 4 = 2 \times 2\]
It can be written as \[ \Rightarrow 4 = {2^2}\]
Now we can factor 9 as \[ \Rightarrow 9 = 3 \times 3\]
It can be written as \[ \Rightarrow 9 = {3^2}\]
Now the given problem as
\[\Rightarrow {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}} = {\left( {\dfrac{{{2^2}}}{{{3^2}}}} \right)^{\dfrac{3}{2}}}\]
Now applying law of brackets we have,
\[\Rightarrow \dfrac{{{{\left( {{2^2}} \right)}^{\dfrac{3}{2}}}}}{{{{\left( {{3^2}} \right)}^{\dfrac{3}{2}}}}}\]
\[\Rightarrow \dfrac{{{2^{2 \times \dfrac{3}{2}}}}}{{{3^{2 \times \dfrac{3}{2}}}}}\]
\[\Rightarrow \dfrac{{{2^3}}}{{{3^3}}}\]
\[\Rightarrow \dfrac{8}{{27}}\].
Thus we have \[ {\left( {\dfrac{4}{9}} \right)^{\dfrac{3}{2}}} = \dfrac{8}{{27}}\]
Additional information:
We know the laws of indices.
Rule 1: if a constant or variable has index as ‘0’, then the result will be equal to one, regardless of any base value. That is \[{a^0} = 1\] and ‘a’ is not equal to zero.
Rule 2: If the index is a negative value, then it can be shown as the reciprocal of the positive index raised to the same variable. That is \[{a^{ - 1}} = \dfrac{1}{a}\].
Rule 3: If we have two variables with the same base and different exponent, we add the exponent that is \[{x^a} \times {x^b} = {x^{a + b}}\].
Rule 4: If we have division of two number with same base, that is \[\dfrac{{{x^a}}}{{{x^b}}} = {x^{a - b}}\].
Rule 5: We also know the bracket law that is \[ {\left( {\dfrac{a}{b}} \right)^n} = \left( {\dfrac{{{a^n}}}{{{b^n}}}} \right)\]
These are some basic indices laws.
Note: These rules or laws of indices help us to minimize the problems and get the answer in very less time. These powers can be positive and negative but can be molded according to our convenience while solving the problem. Also note that cube-root ( to the power \[\dfrac{1}{3}\]), square-root ( to the power \[\dfrac{1}{2}\]) are fractions with 1 as numerator and respective root in denominator but the power in our problem is first cube root then square. For example suppose if we have \[{\left( {25} \right)^{ - \left( {\dfrac{1}{2}} \right)}}\]. We can write it as \[\sqrt {25} \] which gives us the answer 5.
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