
Simplify the above expression.
\[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\]
Answer
615k+ views
Hint: Find the multiplicative inverse of the function. Undergo algebraic operations like multiplication and division, thus simply the given expression.
Complete step-by-step answer:
We have been given the expression, \[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\]
A multiplicative inverse or reciprocal of a number x is denoted by \[{}^{1}/{}_{x}\] or \[{{x}^{-1}}\] is a multiplicative by x yields to the multiplicative identity 1.
The multiplicative inverse of a fraction \[{}^{a}/{}_{b}\] is \[{}^{b}/{}_{a}\]. For the multiplicative inverse of a real number divide 1 by the number.
We have, \[\left( \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}} \right).......(1)\]
The multiplicative inverse of the given real number are,
\[{{3}^{-5}}={}^{1}/{}_{{{3}^{5}}},{{10}^{-5}}={}^{1}/{}_{{{10}^{5}}},{{5}^{-7}}={}^{1}/{}_{{{5}^{7}}},{{6}^{-5}}={}^{1}/{}_{{{6}^{5}}}\]
Substitute these values in equation (1).
\[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}=\dfrac{{}^{1}/{}_{{{3}^{5}}}\times {}^{1}/{}_{{{10}^{5}}}\times 125}{{}^{1}/{}_{{{5}^{7}}}\times {}^{1}/{}_{{{6}^{5}}}}\]
Now this can be written as,
\[\dfrac{{{5}^{7}}\times {{6}^{5}}\times 125}{{{3}^{5}}\times {{10}^{5}}}\]
We can write 125 as \[{{5}^{3}}\]
\[=\dfrac{{{5}^{7}}\times {{6}^{5}}\times {{5}^{3}}}{{{3}^{5}}\times {{10}^{5}}}\]
Now let us simplify the above expression.
\[=\dfrac{{{5}^{10}}\times {{6}^{5}}}{{{3}^{5}}\times {{10}^{5}}}=\dfrac{\left( 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5 \right)\times \left( 6\times 6\times 6\times 6\times 6 \right)}{\left( 3\times 3\times 3\times 3\times 3 \right)\times \left( 10\times 10\times 10\times 10\times 10 \right)}\]
Now cancel out the similar term and their expressions.
\[\begin{align}
& =\dfrac{\left( 5\times 5\times 5\times 5\times 5\times \right)\times \left( 6\times 6\times 6\times 6\times 6 \right)}{\left( 3\times 3\times 3\times 3\times 3 \right)\times \left( 2\times 2\times 2\times 2\times 2 \right)} \\
& =\dfrac{{{5}^{5}}\times \left( {{2}^{5}} \right)}{{{2}^{5}}} \\
\end{align}\]
Cancel out \[{{3}^{5}}\] from numerator to denominator.
\[\begin{align}
& =\dfrac{{{5}^{5}}\times \left( {{2}^{5}}\times {{3}^{5}} \right)}{{{3}^{5}}\times {{2}^{5}}} \\
& ={{5}^{5}}=3125 \\
\end{align}\]
Thus we got the value of \[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}={{5}^{5}}=3125\]
Hence we simplified it.
Note:In the phrase multiplicative inverse, the qualifier multiplicative is often committed and then tacitly understood. Multiplicative inverse can be defined over many mathematical domains as well as number. In the case of real numbers, zero does not have a reciprocal because no real number multiplied by zero produces 1.
Complete step-by-step answer:
We have been given the expression, \[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}\]
A multiplicative inverse or reciprocal of a number x is denoted by \[{}^{1}/{}_{x}\] or \[{{x}^{-1}}\] is a multiplicative by x yields to the multiplicative identity 1.
The multiplicative inverse of a fraction \[{}^{a}/{}_{b}\] is \[{}^{b}/{}_{a}\]. For the multiplicative inverse of a real number divide 1 by the number.
We have, \[\left( \dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}} \right).......(1)\]
The multiplicative inverse of the given real number are,
\[{{3}^{-5}}={}^{1}/{}_{{{3}^{5}}},{{10}^{-5}}={}^{1}/{}_{{{10}^{5}}},{{5}^{-7}}={}^{1}/{}_{{{5}^{7}}},{{6}^{-5}}={}^{1}/{}_{{{6}^{5}}}\]
Substitute these values in equation (1).
\[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}=\dfrac{{}^{1}/{}_{{{3}^{5}}}\times {}^{1}/{}_{{{10}^{5}}}\times 125}{{}^{1}/{}_{{{5}^{7}}}\times {}^{1}/{}_{{{6}^{5}}}}\]
Now this can be written as,
\[\dfrac{{{5}^{7}}\times {{6}^{5}}\times 125}{{{3}^{5}}\times {{10}^{5}}}\]
We can write 125 as \[{{5}^{3}}\]
\[=\dfrac{{{5}^{7}}\times {{6}^{5}}\times {{5}^{3}}}{{{3}^{5}}\times {{10}^{5}}}\]
Now let us simplify the above expression.
\[=\dfrac{{{5}^{10}}\times {{6}^{5}}}{{{3}^{5}}\times {{10}^{5}}}=\dfrac{\left( 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5\times 5 \right)\times \left( 6\times 6\times 6\times 6\times 6 \right)}{\left( 3\times 3\times 3\times 3\times 3 \right)\times \left( 10\times 10\times 10\times 10\times 10 \right)}\]
Now cancel out the similar term and their expressions.
\[\begin{align}
& =\dfrac{\left( 5\times 5\times 5\times 5\times 5\times \right)\times \left( 6\times 6\times 6\times 6\times 6 \right)}{\left( 3\times 3\times 3\times 3\times 3 \right)\times \left( 2\times 2\times 2\times 2\times 2 \right)} \\
& =\dfrac{{{5}^{5}}\times \left( {{2}^{5}} \right)}{{{2}^{5}}} \\
\end{align}\]
Cancel out \[{{3}^{5}}\] from numerator to denominator.
\[\begin{align}
& =\dfrac{{{5}^{5}}\times \left( {{2}^{5}}\times {{3}^{5}} \right)}{{{3}^{5}}\times {{2}^{5}}} \\
& ={{5}^{5}}=3125 \\
\end{align}\]
Thus we got the value of \[\dfrac{{{3}^{-5}}\times {{10}^{-5}}\times 125}{{{5}^{-7}}\times {{6}^{-5}}}={{5}^{5}}=3125\]
Hence we simplified it.
Note:In the phrase multiplicative inverse, the qualifier multiplicative is often committed and then tacitly understood. Multiplicative inverse can be defined over many mathematical domains as well as number. In the case of real numbers, zero does not have a reciprocal because no real number multiplied by zero produces 1.
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