Question

Simplify the given problem using the laws of exponents: $\dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$.

Answer 1

Hint: We start solving the problem by converting all the numbers in both numerator and denominator into the exponential form. We then use the fact ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ in both numerator and denominator to proceed through the problem. We then use the facts $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ if $m>n$ and $\dfrac{{{a}^{m}}}{{{a}^{n}}}=\dfrac{1}{{{a}^{n-m}}}$ if m < n and make subsequent calculations to get the required value.

Complete step-by-step answer:
According the problem, we need to find the value of $\dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$ using the laws of exponents.
We have $\dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$ ---(1).
We need to convert all the numbers that were given in the equations in exponential in order to use the laws of exponents. From equation (1), we can see 81 and 125 are not in exponential form. So, we convert them.
We know that $81={{3}^{4}}$ and $125={{5}^{3}}$. We use these results in equation (1).
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{4}}\times {{3}^{8}}\times {{5}^{-5}}}{{{5}^{3}}\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$---(2).
We know that ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We use this result in equation (2). We use this for the powers of 3 and 5 in both numerator and denominator.
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{4+8}}\times {{5}^{-5}}}{{{5}^{3+9}}\times {{3}^{-6+4}}}$.
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{12}}\times {{5}^{-5}}}{{{5}^{12}}\times {{3}^{-2}}}$ ---(3).
We know that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ if m > n and $\dfrac{{{a}^{m}}}{{{a}^{n}}}=\dfrac{1}{{{a}^{n-m}}}$ if $m < n$. We use this results in equation (3). We can see that the power of 3 in denominator is less than the power of 3 in numerator. We can also see that the power of 3 in denominator is greater than the power of 5 in numerator.
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{12-\left( -2 \right)}}}{{{5}^{12-\left( -5 \right)}}}$.
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{12+2}}}{{{5}^{12+5}}}$.
$\Rightarrow \dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}=\dfrac{{{3}^{14}}}{{{5}^{17}}}$.
We have found the value of $\dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$ as $\dfrac{{{3}^{14}}}{{{5}^{17}}}$.
∴ The value of $\dfrac{81\times {{3}^{8}}\times {{5}^{-5}}}{125\times {{3}^{-6}}\times {{3}^{4}}\times {{5}^{9}}}$ is $\dfrac{{{3}^{14}}}{{{5}^{17}}}$.

Note: We should not always use the fact $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ as it is only true for $m>n$. We can see from the final answer all the elements in both numerator and denominator are in terms of powers. We should not make any mistakes while performing addition and subtraction operations. Similarly, we can expect the problems that requires to use the following facts: ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ and also the equations that contain equalities to find the value of certain variable.