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Simplify the given expression.
$\dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 4}}}}\left( {t \ne 0} \right)$

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Last updated date: 16th Jul 2024
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Answer
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Hint – In this question we have to simplify the given equation so look for terms common in both numerator and denominator part, use the basic property of exponential powers like if the base is the same and the number is in multiplication therefore powers get added to get the answer.

Complete step-by-step answer:
Given equation is
$\dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 4}}}}\left( {t \ne 0} \right)$
Now we have to simplify this equation.
First of all cancel out the common terms from numerator and denominator.
So, as we see that $\left( {{t^{ - 4}}} \right)$ is common so cancel out this term we have,
$ \Rightarrow \dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 4}}}} = \dfrac{{25}}{{{5^{ - 3}} \times 10}}$ So, this is the remaining term we have
Now factorize 25 and 10 we have,
$ \Rightarrow \dfrac{{25}}{{{5^{ - 3}} \times 10}} = \dfrac{{5 \times 5}}{{{5^{ - 3}} \times 2 \times 5}}$
So, as we see that 5 is common so cancel out this term we have,
$ \Rightarrow \dfrac{{5 \times 5}}{{{5^{ - 3}} \times 2 \times 5}} = \dfrac{5}{{2 \times {5^{ - 3}}}}$
Now take $\left( {{5^{ - 3}}} \right)$ to numerator we have,
$ \Rightarrow \dfrac{5}{{2 \times {5^{ - 3}}}} = \dfrac{{5 \times {5^3}}}{2}$
Now if the base is the same and the number is in multiplication therefore powers got added.
$ \Rightarrow \dfrac{{5 \times {5^3}}}{2} = \dfrac{{{5^{1 + 3}}}}{2} = \dfrac{{{5^4}}}{2}$
Now as we know ${5^4} = 625$ so, substitute this value in above equation we have,
$ \Rightarrow \dfrac{{25 \times {t^{ - 4}}}}{{{5^{ - 3}} \times 10 \times {t^{ - 4}}}} = \dfrac{{625}}{2} = 312.5$
So, this is the required answer.

Note – Whenever we face such type of problems the key concept is to use basic simplification techniques to get maximum terms canceled, in this problem it’s given that $t \ne 0$, because if we make t zero then this equation will have$\dfrac{0}{0}$, indeterminate form which can’t be solved thus take care of the conditions.