Question

Simplify $\dfrac{{{15}^{3}}\times {{18}^{2}}}{{{3}^{2}}\times {{5}^{4}}\times {{12}^{2}}}$ \[\]

Answer 1

Hint: We first prime factorize the composite numbers in both numerator and denominator and then use the formula ${{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$. We arrange them in ascending order of the bases and use the formula ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ . We make fractions with same base at numerator and denominator and use the formula $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. We finally get the result using ${{a}^{-1}}=\dfrac{1}{a}$.\[\]

Complete step by step answer:
The given expression is ,
\[\dfrac{{{15}^{3}}\times {{18}^{2}}}{{{3}^{2}}\times {{5}^{4}}\times {{12}^{2}}}\]
We see that there are total five numbers in the expression with exponents as bases out of which 3 and 5 are prime numbers the rest three 15,18,12 are composite numbers.
We first prime factorize the composite numbers 15,18,12 and get
\[\begin{align}
  & 15=3\times 5 \\
 & 18=2\times 3\times 3 \\
 & 12=2\times 2\times 3 \\
\end{align}\]

Now we shall replace the given composite numbers 15, 18, 12 with their prime factorization to get ,
\[\dfrac{{{15}^{3}}\times {{18}^{2}}}{{{3}^{2}}\times {{5}^{4}}\times {{12}^{2}}}=\dfrac{{{\left( 3\times 5 \right)}^{3}}\times {{\left( 2\times 3\times 3 \right)}^{2}}}{{{3}^{2}}\times {{5}^{4}}\times {{\left( 2\times 2\times 3 \right)}^{2}}}\]

We know that two numbers say $a$ and $b$ with exponents $m$ then the exponent of the product ${{\left( a\times b \right)}^{m}}={{a}^{m}}\times {{b}^{m}}$ . W use it in the above result and get,
\[=\dfrac{{{3}^{3}}\times {{5}^{3}}\times {{2}^{2}}\times {{3}^{2}}\times {{3}^{2}}}{{{3}^{2}}\times {{5}^{4}}\times {{2}^{2}}\times {{2}^{2}}\times {{3}^{2}}}\]
We rearrange the order of multiplication in numerator and then in denominator such that the terms in ascending order of the base so that terms with same base will be next to each other. We can do it because multiplication is commutative.
\[=\dfrac{{{2}^{2}}\times {{3}^{3}}\times {{3}^{2}}\times {{3}^{2}}\times {{5}^{3}}}{{{2}^{2}}\times {{2}^{2}}\times {{3}^{2}}\times {{3}^{2}}\times {{5}^{4}}}\]
We know that when we multiply two numbers with same base their exponents are added up. We write it in symbols for real number $a$ with exponents $m,n$ respectively then ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We use it in the above result and get ,
\[\begin{align}
  & =\dfrac{{{2}^{2}}\times {{3}^{3+2+2}}\times {{5}^{3}}}{{{2}^{2+2}}\times {{3}^{2+2}}\times {{5}^{4}}} \\
 & =\dfrac{{{2}^{2}}\times {{3}^{7}}\times {{5}^{3}}}{{{2}^{4}}\times {{3}^{4}}\times {{5}^{4}}} \\
 & =\dfrac{{{2}^{2}}}{{{2}^{4}}}\times \dfrac{{{3}^{7}}}{{{3}^{3}}}\times \dfrac{{{5}^{3}}}{{{5}^{4}}} \\
\end{align}\]
We know that when we divide two numbers with the same base the exponent of the denominator is subtracted from the exponent of the numerator. We write it in symbols for real number $a$ with exponents $m,n$ respectively then $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$. We use it in the above result and get
\[\begin{align}
  & ={{2}^{2-4}}\times {{3}^{7-3}}\times {{5}^{3-4}} \\
 & ={{2}^{-2}}\times {{3}^{4}}\times {{5}^{-1}} \\
\end{align}\]
We know for any non-zero real number $a$, ${{a}^{-1}}=\dfrac{1}{a}$. So we have,
\[\begin{align}
  & ={{2}^{-2}}\times {{3}^{4}}\times {{5}^{-1}} \\
 & =\dfrac{1}{4}\times 81\times \dfrac{1}{5} \\
 & =\dfrac{81}{20} \\
\end{align}\]

Note: When we are dealing with power and exponents we need to be careful that both base and the exponent not be zero at the same time. We also note that in ${{a}^{-1}}=\dfrac{1}{a}$, $a$ cannot be zero and ${{a}^{-1}}$ is also called the multiplicative inverse or the reciprocal of $a$.