Question

Evaluate \[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\].

Answer 1

Hint: We need to evaluate \[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\].. We will use the properties, \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\] , \[a \div b = \dfrac{a}{b}\], \[a \div b = a \times \dfrac{1}{b}\]and \[\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}\]. Using these properties, we will first simplify the numerator and then denominator. After individually simplifying numerator and denominator, we will write them as a fraction and then simplify further.

Complete step by step answer:
We need to simplify \[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\].
First simplifying the numerator \[{8^{ - 1}} \times {5^3}\] using\[{a^{ - n}} = \dfrac{1}{{{a^n}}}\], we get
\[ \Rightarrow {8^{ - 1}} \times {5^3} = \dfrac{1}{8} \times {5^3}\]
Also, \[{a^3} = a \times a \times a\]. Using this, we have
\[ \Rightarrow {8^{ - 1}} \times {5^3} = \dfrac{1}{8} \times 5 \times 5 \times 5\]
Writing \[a = \dfrac{a}{1}\], we get
\[ \Rightarrow {8^{ - 1}} \times {5^3} = \dfrac{1}{8} \times 5 \times 5 \times 5 = \dfrac{1}{8} \times \dfrac{5}{1} \times \dfrac{5}{1} \times \dfrac{5}{1}\]
Now, Solving numerator and denominator
\[ \Rightarrow {8^{ - 1}} \times {5^3} = \dfrac{1}{8} \times 5 \times 5 \times 5 = \dfrac{1}{8} \times \dfrac{5}{1} \times \dfrac{5}{1} \times \dfrac{5}{1}\]
\[\Rightarrow {8^{ - 1}} \times {5^3} = \dfrac{{125}}{8}\]
Hence, we got numerator \[{8^{ - 1}} \times {5^3} = \dfrac{{125}}{8} - - - - - - (1)\]

Now solving the denominator \[{2^{ - 4}}\] using \[{a^{ - n}} = \dfrac{1}{{{a^n}}}\], we get
\[{2^{ - 4}} = \dfrac{1}{{{2^4}}}\]
Now using \[{a^n} = \underbrace {a \times a \times a \times ....... \times a}_{(n)times}\], we have
\[ \Rightarrow \dfrac{1}{{{2^4}}} = \dfrac{1}{{2 \times 2 \times 2 \times 2}}\]
\[\Rightarrow \dfrac{1}{{{2^4}}} = \dfrac{1}{{16}}\]
Hence, we got denominator \[{2^{ - 4}} = \dfrac{1}{{16}} - - - - - - (2)\]
Now, from (1) and (2)
\[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{\dfrac{{125}}{8}}}{{\dfrac{1}{{16}}}}\]
Using \[\dfrac{a}{b} = a \div b\], we have
\[ \Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{\dfrac{{125}}{8}}}{{\dfrac{1}{{16}}}} \\
\Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}= \dfrac{{125}}{8} \div \dfrac{1}{{16}}\]

Now using \[a \div b = a \times \dfrac{1}{b}\] and \[\dfrac{1}{{\dfrac{a}{b}}} = \dfrac{b}{a}\], we have
\[ \Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{\dfrac{{125}}{8}}}{{\dfrac{1}{{16}}}} \\
\Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}= \dfrac{{125}}{8} \div \dfrac{1}{{16}}\]
\[\Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{125}}{8} \times \dfrac{1}{{\dfrac{1}{{16}}}}\]
\[ \Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}= \dfrac{{125}}{8} \times \dfrac{{16}}{1}\]
After cancelling out the terms from numerator and denominator, we have
\[ \Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{125}}{1} \times \dfrac{2}{1}\]
Solving the numerator and denominator, we have
\[ \Rightarrow \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = \dfrac{{125 \times 2}}{{1 \times 1}} = \dfrac{{250}}{1}\]
\[ \therefore \dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}} = 250\]

Hence, the value of \[\dfrac{{{8^{ - 1}} \times {5^3}}}{{{2^{ - 4}}}}\] is 250.

Note: We could have solved the numerator and denominator simultaneously also. When we solve fractions in fraction, we usually get confused with the terms that whether they should be in numerator or denominator. We need to be careful while we are solving and so its better to write the fraction $\dfrac{a}{b}$ as $a\div b$ and then solve. While we are cancelling the terms, we should keep in mind that the term is cancelled only once and not twice.