Find the number of prime factors in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$.
A. 6
B. 7
C. 8
D. none of these
Answer
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Hint: We need to use the law of indices to simplify the problem. There are 3 concepts of indices which we will use. We will convert the individual terms in the form of factorisation and their powers. Then combine the similar base terms and calculate their power. At the end we find the number of prime factors in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$.
Complete step-by-step answer:
We have been given the multiplications of three terms in indices form.
We will try to form the equation in its simplest form by using the laws of indices.
Also, we need to break them to the power of their prime factors.
We know that ${{a}^{\dfrac{1}{m}}}=\sqrt[m]{a},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}},{{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
There are three terms in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$ which we will break individually.
The first term is ${{\left( 216 \right)}^{\dfrac{3}{5}}}$. Using the laws of indices, we get
${{\left( 216 \right)}^{\dfrac{3}{5}}}={{\left[ {{\left( 216 \right)}^{\dfrac{1}{5}}} \right]}^{^{3}}}$. We know that $216={{2}^{3}}\times {{3}^{3}}$.
So, \[{{\left( 216 \right)}^{\dfrac{3}{5}}}={{\left( {{2}^{3}}\times {{3}^{3}} \right)}^{\dfrac{3}{5}}}={{2}^{3\times \dfrac{3}{5}}}\times {{3}^{3\times \dfrac{3}{5}}}={{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}}\]
The second term is ${{\left( 2500 \right)}^{\dfrac{2}{5}}}$. Using the laws of indices, we get
${{\left( 2500 \right)}^{\dfrac{2}{5}}}={{\left[ {{\left( 2500 \right)}^{\dfrac{1}{5}}} \right]}^{^{2}}}$. We know that $2500={{2}^{2}}\times {{5}^{4}}$.
So, \[{{\left( 2500 \right)}^{\dfrac{2}{5}}}={{\left( {{2}^{2}}\times {{5}^{4}} \right)}^{\dfrac{2}{5}}}={{2}^{2\times \dfrac{2}{5}}}\times {{5}^{4\times \dfrac{2}{5}}}={{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}}\]
The third term is ${{\left( 300 \right)}^{\dfrac{1}{5}}}$. We know that $300={{2}^{2}}\times 3\times {{5}^{2}}$.
Using the laws of indices, we get
So, \[{{\left( 300 \right)}^{\dfrac{1}{5}}}={{\left( {{2}^{2}}\times 3\times {{5}^{2}} \right)}^{\dfrac{1}{5}}}={{2}^{2\times \dfrac{1}{5}}}\times {{3}^{1\times \dfrac{1}{5}}}\times {{5}^{2\times \dfrac{1}{5}}}={{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}}\]
Now we need to multiply all of them
${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}=\left( {{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}} \right)\times \left( {{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}} \right)\times \left( {{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}} \right)$
We take the similar base terms together and find the solution
\[\begin{align}
& \left( {{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}} \right)\times \left( {{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}} \right)\times \left( {{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}} \right) \\
& =\left( {{2}^{\dfrac{9}{5}}}\times {{2}^{\dfrac{4}{5}}}\times {{2}^{\dfrac{2}{5}}} \right)\times \left( {{3}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{1}{5}}} \right)\times \left( {{5}^{\dfrac{8}{5}}}\times {{5}^{\dfrac{2}{5}}} \right) \\
& ={{2}^{\dfrac{9}{5}+\dfrac{4}{5}+\dfrac{2}{5}}}\times {{3}^{\dfrac{9}{5}+\dfrac{1}{5}}}\times {{5}^{\dfrac{8}{5}+\dfrac{2}{5}}} \\
& ={{2}^{\dfrac{9+4+2}{5}}}\times {{3}^{\dfrac{9+1}{5}}}\times {{5}^{\dfrac{8+2}{5}}} \\
& ={{2}^{\dfrac{15}{5}}}\times {{3}^{\dfrac{10}{5}}}\times {{5}^{\dfrac{10}{5}}} \\
& ={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}} \\
\end{align}\]
So, we expressed ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}$ in its simpler form.
We found there are three prime factors in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$ which are 2, 3, 5.
So, the correct answer is “Option D”.
Note: We need to be careful about the laws. We need to remember the difference between ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and ${{a}^{{{m}^{n}}}}$ which are different to each other. Also, we need to find the prime factors not the actual value of the equation. So, keeping it in the form of multiplication is necessary.
Complete step-by-step answer:
We have been given the multiplications of three terms in indices form.
We will try to form the equation in its simplest form by using the laws of indices.
Also, we need to break them to the power of their prime factors.
We know that ${{a}^{\dfrac{1}{m}}}=\sqrt[m]{a},{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}},{{a}^{m}}.{{a}^{n}}={{a}^{m+n}},\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
There are three terms in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$ which we will break individually.
The first term is ${{\left( 216 \right)}^{\dfrac{3}{5}}}$. Using the laws of indices, we get
${{\left( 216 \right)}^{\dfrac{3}{5}}}={{\left[ {{\left( 216 \right)}^{\dfrac{1}{5}}} \right]}^{^{3}}}$. We know that $216={{2}^{3}}\times {{3}^{3}}$.
So, \[{{\left( 216 \right)}^{\dfrac{3}{5}}}={{\left( {{2}^{3}}\times {{3}^{3}} \right)}^{\dfrac{3}{5}}}={{2}^{3\times \dfrac{3}{5}}}\times {{3}^{3\times \dfrac{3}{5}}}={{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}}\]
The second term is ${{\left( 2500 \right)}^{\dfrac{2}{5}}}$. Using the laws of indices, we get
${{\left( 2500 \right)}^{\dfrac{2}{5}}}={{\left[ {{\left( 2500 \right)}^{\dfrac{1}{5}}} \right]}^{^{2}}}$. We know that $2500={{2}^{2}}\times {{5}^{4}}$.
So, \[{{\left( 2500 \right)}^{\dfrac{2}{5}}}={{\left( {{2}^{2}}\times {{5}^{4}} \right)}^{\dfrac{2}{5}}}={{2}^{2\times \dfrac{2}{5}}}\times {{5}^{4\times \dfrac{2}{5}}}={{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}}\]
The third term is ${{\left( 300 \right)}^{\dfrac{1}{5}}}$. We know that $300={{2}^{2}}\times 3\times {{5}^{2}}$.
Using the laws of indices, we get
So, \[{{\left( 300 \right)}^{\dfrac{1}{5}}}={{\left( {{2}^{2}}\times 3\times {{5}^{2}} \right)}^{\dfrac{1}{5}}}={{2}^{2\times \dfrac{1}{5}}}\times {{3}^{1\times \dfrac{1}{5}}}\times {{5}^{2\times \dfrac{1}{5}}}={{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}}\]
Now we need to multiply all of them
${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}=\left( {{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}} \right)\times \left( {{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}} \right)\times \left( {{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}} \right)$
We take the similar base terms together and find the solution
\[\begin{align}
& \left( {{2}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{9}{5}}} \right)\times \left( {{2}^{\dfrac{4}{5}}}\times {{5}^{\dfrac{8}{5}}} \right)\times \left( {{2}^{\dfrac{2}{5}}}\times {{3}^{\dfrac{1}{5}}}\times {{5}^{\dfrac{2}{5}}} \right) \\
& =\left( {{2}^{\dfrac{9}{5}}}\times {{2}^{\dfrac{4}{5}}}\times {{2}^{\dfrac{2}{5}}} \right)\times \left( {{3}^{\dfrac{9}{5}}}\times {{3}^{\dfrac{1}{5}}} \right)\times \left( {{5}^{\dfrac{8}{5}}}\times {{5}^{\dfrac{2}{5}}} \right) \\
& ={{2}^{\dfrac{9}{5}+\dfrac{4}{5}+\dfrac{2}{5}}}\times {{3}^{\dfrac{9}{5}+\dfrac{1}{5}}}\times {{5}^{\dfrac{8}{5}+\dfrac{2}{5}}} \\
& ={{2}^{\dfrac{9+4+2}{5}}}\times {{3}^{\dfrac{9+1}{5}}}\times {{5}^{\dfrac{8+2}{5}}} \\
& ={{2}^{\dfrac{15}{5}}}\times {{3}^{\dfrac{10}{5}}}\times {{5}^{\dfrac{10}{5}}} \\
& ={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}} \\
\end{align}\]
So, we expressed ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}={{2}^{3}}\times {{3}^{2}}\times {{5}^{2}}$ in its simpler form.
We found there are three prime factors in ${{\left( 216 \right)}^{\dfrac{3}{5}}}\times {{\left( 2500 \right)}^{\dfrac{2}{5}}}\times {{\left( 300 \right)}^{\dfrac{1}{5}}}$ which are 2, 3, 5.
So, the correct answer is “Option D”.
Note: We need to be careful about the laws. We need to remember the difference between ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and ${{a}^{{{m}^{n}}}}$ which are different to each other. Also, we need to find the prime factors not the actual value of the equation. So, keeping it in the form of multiplication is necessary.
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