
How do you simplify \[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}}\] ?
Answer
556.5k+ views
Hint: If we have a fraction raised to a negative index we can make the index as positive by taking reciprocal of the fraction. That if we have \[{\left( {\dfrac{a}{b}} \right)^{ - n}}\] , this can be written as \[{\left( {\dfrac{b}{a}} \right)^n}\] . Here ‘n’ is natural numbers or fraction. In the given problem we gave a fraction as an index. After that we can express the numerator and denominator number as power of a number. Simplifying we will have a desired result.
Complete step-by-step answer:
Given,
\[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}}\]
This can be written as \[{\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}}\]
To express 81 as a power of a number, let’s find the factors of 81.
The factors of 81 are 3, 3, 3 and 3. That is \[81 = 3 \times 3 \times 3 \times 3\]
Therefore \[81 = {3^4}\] .
To express 16 as a power of a number, let’s find the factors of 16.
The factors of 16 are 3, 2, 2 and 2. That is \[16 = 2 \times 2 \times 2 \times 2\]
Therefore \[16 = {2^4}\] .
Now we have,
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} = {\left( {\dfrac{{{3^4}}}{{{2^4}}}} \right)^{\dfrac{3}{4}}}\]
This can be written as
\[ = \dfrac{{{{\left( {{3^4}} \right)}^{^{\dfrac{3}{4}}}}}}{{{{\left( {{2^4}} \right)}^{^{\dfrac{3}{4}}}}}}\]
We know the rule of brackets in the law of indices. That is \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] .
\[ = \dfrac{{{{\left( 3 \right)}^{4 \times \dfrac{3}{4}}}}}{{{{\left( 2 \right)}^{4 \times \dfrac{3}{4}}}}}\]
\[ = \dfrac{{{3^3}}}{{{2^3}}}\]
\[ = \dfrac{{27}}{8}\]
Hence we have, \[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}} = \dfrac{{27}}{8}\] . This is the exact form.
We can put the required answer in decimal form also,
That is,
\[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}} = 3.375\] . This is in the decimal form.
So, the correct answer is “$\dfrac{{27}}{8}$”.
Note: In this type of problem we express the numerator and the denominator numbers as powers of some number. By doing this we can simplify the given problem easily. Above all we did is simple multiplication and division. We use the law of indices to simplify the given problem. Careful in the calculation part.
Complete step-by-step answer:
Given,
\[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}}\]
This can be written as \[{\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}}\]
To express 81 as a power of a number, let’s find the factors of 81.
The factors of 81 are 3, 3, 3 and 3. That is \[81 = 3 \times 3 \times 3 \times 3\]
Therefore \[81 = {3^4}\] .
To express 16 as a power of a number, let’s find the factors of 16.
The factors of 16 are 3, 2, 2 and 2. That is \[16 = 2 \times 2 \times 2 \times 2\]
Therefore \[16 = {2^4}\] .
Now we have,
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{3}{4}}} = {\left( {\dfrac{{{3^4}}}{{{2^4}}}} \right)^{\dfrac{3}{4}}}\]
This can be written as
\[ = \dfrac{{{{\left( {{3^4}} \right)}^{^{\dfrac{3}{4}}}}}}{{{{\left( {{2^4}} \right)}^{^{\dfrac{3}{4}}}}}}\]
We know the rule of brackets in the law of indices. That is \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\] .
\[ = \dfrac{{{{\left( 3 \right)}^{4 \times \dfrac{3}{4}}}}}{{{{\left( 2 \right)}^{4 \times \dfrac{3}{4}}}}}\]
\[ = \dfrac{{{3^3}}}{{{2^3}}}\]
\[ = \dfrac{{27}}{8}\]
Hence we have, \[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}} = \dfrac{{27}}{8}\] . This is the exact form.
We can put the required answer in decimal form also,
That is,
\[{\left( {\dfrac{{16}}{{81}}} \right)^{ - \dfrac{3}{4}}} = 3.375\] . This is in the decimal form.
So, the correct answer is “$\dfrac{{27}}{8}$”.
Note: In this type of problem we express the numerator and the denominator numbers as powers of some number. By doing this we can simplify the given problem easily. Above all we did is simple multiplication and division. We use the law of indices to simplify the given problem. Careful in the calculation part.
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