Answer
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Hint: We can see that the terms in the expression can be expressed as powers of 2, 5 and 7. In the given expression replace 25 by \[{{5}^{2}}\] , 343 by \[{{7}^{3}}\] , 16 by \[{{2}^{4}}\] and 8 by \[{{2}^{3}}\] . Then solve the given expression using formula \[{{({{x}^{m}})}^{n}}={{x}^{mn}}\] , \[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\] , and \[{{x}^{m}}\times \dfrac{1}{{{x}^{n}}}={{x}^{m-n}}\] .
Complete step-by-step solution -
According to the question, we have the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] .
We can write 25 as a square of 5.
\[{{5}^{2}}=25\] …………(1)
Similarly, we can write 343 as the cube of 7.
\[{{7}^{3}}=343\] ………………..(2)
Similarly, we can write 8 as a cube of 2.
\[{{2}^{3}}=8\] ………………….(3)
Similarly, we can write 16 as $4^{th}$ power of 2.
\[{{2}^{4}}=16\] ………………….(4)
Using equation (1), equation (2), equation (3),and equation (4), we can replace 25, 343, 8 and 16 in the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] .
Now, replacing 25 by \[{{5}^{2}}\] , 343 by \[{{7}^{3}}\] , 16 by \[{{2}^{4}}\] and 8 by \[{{2}^{3}}\] in the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] , we get
\[\dfrac{{{({{5}^{2}})}^{\dfrac{3}{2}}}\times {{({{7}^{3}})}^{\dfrac{3}{5}}}}{{{({{2}^{4}})}^{\dfrac{5}{4}}}\times {{({{2}^{3}})}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] ……………….(5)
Solving equation (5), using the formula \[{{({{x}^{m}})}^{n}}={{x}^{mn}}\] , we get
\[\dfrac{{{({{5}^{2}})}^{\dfrac{3}{2}}}\times {{({{7}^{3}})}^{\dfrac{3}{5}}}}{{{({{2}^{4}})}^{\dfrac{5}{4}}}\times {{({{2}^{3}})}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{\dfrac{6}{2}}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{\dfrac{20}{4}}}\times {{(2)}^{\dfrac{12}{3}}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{5}}\times {{(2)}^{4}}\times {{7}^{\dfrac{3}{5}}}}\] …………………(6)
Now, simplifying equation (6), using the formula \[{{x}^{m}}\times \dfrac{1}{{{x}^{n}}}={{x}^{m-n}}\] and \[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\] , we get
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{5}}\times {{(2)}^{4}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}-{\dfrac{3}{5}}}}}{{{(2)}^{5+{4}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{6}{5}}}}{{{(2)}^{9}}}\]
Hence, the value of the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] is \[\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{6}{5}}}}{{{(2)}^{9}}}\].
Note: In this question, one can think to put the numerical values of each exponential term in the expression given. But, if we do so then our calculations will be lengthy and we should also know the value of each exponential term which is not easy. So, whenever one has to solve this type of question, one should try to minimize the calculation by expressing the terms as powers of smallest possible factors like 2, 3, 5, 7,... and avoid any silly mistakes.
Complete step-by-step solution -
According to the question, we have the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] .
We can write 25 as a square of 5.
\[{{5}^{2}}=25\] …………(1)
Similarly, we can write 343 as the cube of 7.
\[{{7}^{3}}=343\] ………………..(2)
Similarly, we can write 8 as a cube of 2.
\[{{2}^{3}}=8\] ………………….(3)
Similarly, we can write 16 as $4^{th}$ power of 2.
\[{{2}^{4}}=16\] ………………….(4)
Using equation (1), equation (2), equation (3),and equation (4), we can replace 25, 343, 8 and 16 in the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] .
Now, replacing 25 by \[{{5}^{2}}\] , 343 by \[{{7}^{3}}\] , 16 by \[{{2}^{4}}\] and 8 by \[{{2}^{3}}\] in the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] , we get
\[\dfrac{{{({{5}^{2}})}^{\dfrac{3}{2}}}\times {{({{7}^{3}})}^{\dfrac{3}{5}}}}{{{({{2}^{4}})}^{\dfrac{5}{4}}}\times {{({{2}^{3}})}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] ……………….(5)
Solving equation (5), using the formula \[{{({{x}^{m}})}^{n}}={{x}^{mn}}\] , we get
\[\dfrac{{{({{5}^{2}})}^{\dfrac{3}{2}}}\times {{({{7}^{3}})}^{\dfrac{3}{5}}}}{{{({{2}^{4}})}^{\dfrac{5}{4}}}\times {{({{2}^{3}})}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{\dfrac{6}{2}}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{\dfrac{20}{4}}}\times {{(2)}^{\dfrac{12}{3}}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{5}}\times {{(2)}^{4}}\times {{7}^{\dfrac{3}{5}}}}\] …………………(6)
Now, simplifying equation (6), using the formula \[{{x}^{m}}\times \dfrac{1}{{{x}^{n}}}={{x}^{m-n}}\] and \[{{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}\] , we get
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}}}}{{{(2)}^{5}}\times {{(2)}^{4}}\times {{7}^{\dfrac{3}{5}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{9}{5}-{\dfrac{3}{5}}}}}{{{(2)}^{5+{4}}}}\]
\[=\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{6}{5}}}}{{{(2)}^{9}}}\]
Hence, the value of the expression \[\dfrac{{{25}^{\dfrac{3}{2}}}\times {{343}^{\dfrac{3}{5}}}}{{{16}^{\dfrac{5}{4}}}\times {{8}^{\dfrac{4}{3}}}\times {{7}^{\dfrac{3}{5}}}}\] is \[\dfrac{{{(5)}^{3}}\times {{(7)}^{\dfrac{6}{5}}}}{{{(2)}^{9}}}\].
Note: In this question, one can think to put the numerical values of each exponential term in the expression given. But, if we do so then our calculations will be lengthy and we should also know the value of each exponential term which is not easy. So, whenever one has to solve this type of question, one should try to minimize the calculation by expressing the terms as powers of smallest possible factors like 2, 3, 5, 7,... and avoid any silly mistakes.
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