Answer
451.8k+ views
Hint: Direct division can be used as an early approach to this type of problem to cross-check the answers. Also, it can be used to ease the calculations for numeric parts which are easily cancelled out through divisions and variables are transposed and solved using properties of exponents.
Complete step-by-step answer:
Here, we have the given equation as $\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}\left( t\ne 0 \right)$.
We have to apply most of the properties of exponents in the above given equation, i.e.,
$\begin{align}
& \Rightarrow {{a}^{x}}=\dfrac{1}{{{a}^{-x}}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{b}^{y}}}={{a}^{x}}\times {{b}^{-y}} \\
& \Rightarrow {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}...\text{ }\left( 1 \right) \\
\end{align}$
Thus, from the given equation, we have
$=\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}\left( t\ne 0 \right)$
Transposing values from denominator to numerator, we get
\[\begin{align}
& =\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}} \\
& =\dfrac{25\times {{5}^{-\left( -2 \right)}}\times {{t}^{-4}}\times {{t}^{-\left( -8 \right)}}}{10} \\
& =\dfrac{25\times {{5}^{2}}\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{25\times 25\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{625\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
\end{align}\]
Now, applying another property of exponent from equation (1), we get
$\begin{align}
& =\dfrac{625\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{625\times {{t}^{-4+8}}}{10} \\
& =\dfrac{625\times {{t}^{4}}}{10} \\
\end{align}$
Applying simple division on numeric terms, we get
$\begin{align}
& =\dfrac{625\times {{t}^{4}}}{10} \\
& =62.5{{t}^{4}} \\
\end{align}$
Hence, the given equation can be simplified to $62.5{{t}^{4}}$.
Note: This type of problems can be easily solved by using direct division or using the properties of exponents. An error which can be made here is that when exponents are transposed then there can be changes in sign conventions in their powers, which needs to be kept in mind.
Complete step-by-step answer:
Here, we have the given equation as $\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}\left( t\ne 0 \right)$.
We have to apply most of the properties of exponents in the above given equation, i.e.,
$\begin{align}
& \Rightarrow {{a}^{x}}=\dfrac{1}{{{a}^{-x}}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{b}^{y}}}={{a}^{x}}\times {{b}^{-y}} \\
& \Rightarrow {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}}...\text{ }\left( 1 \right) \\
\end{align}$
Thus, from the given equation, we have
$=\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}}\left( t\ne 0 \right)$
Transposing values from denominator to numerator, we get
\[\begin{align}
& =\dfrac{25\times {{t}^{-4}}}{{{5}^{-2}}\times 10\times {{t}^{-8}}} \\
& =\dfrac{25\times {{5}^{-\left( -2 \right)}}\times {{t}^{-4}}\times {{t}^{-\left( -8 \right)}}}{10} \\
& =\dfrac{25\times {{5}^{2}}\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{25\times 25\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{625\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
\end{align}\]
Now, applying another property of exponent from equation (1), we get
$\begin{align}
& =\dfrac{625\times {{t}^{-4}}\times {{t}^{8}}}{10} \\
& =\dfrac{625\times {{t}^{-4+8}}}{10} \\
& =\dfrac{625\times {{t}^{4}}}{10} \\
\end{align}$
Applying simple division on numeric terms, we get
$\begin{align}
& =\dfrac{625\times {{t}^{4}}}{10} \\
& =62.5{{t}^{4}} \\
\end{align}$
Hence, the given equation can be simplified to $62.5{{t}^{4}}$.
Note: This type of problems can be easily solved by using direct division or using the properties of exponents. An error which can be made here is that when exponents are transposed then there can be changes in sign conventions in their powers, which needs to be kept in mind.
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