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# How do you simplify ${\left( {\dfrac{{25}}{{16}}} \right)^{ - \dfrac{3}{2}}}$?

Last updated date: 01st Mar 2024
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Hint: In order to determine the simplest form of the above question ,first take the reciprocal to remove negative sign from the exponent and then write $16 = {4^2}\,and\,25 = {5^2}$ and divide the exponents using identity ${\left( {{a^m}} \right)^{\dfrac{1}{n}}} = {a^{\dfrac{m}{n}}}$ to get your required simplification.

Complete step by step solution:
We are given a rational number ${\left( {\dfrac{{25}}{{16}}} \right)^{ - \dfrac{3}{2}}}$.
In order to simplify the above rational number we will be using some of the exponential identities ,to rewrite the rational number and simplify it.
Since, any number raised to negative power , is actually the reciprocal of it .
Taking the reciprocal of our given function to remove the negative sign from the exponent value .
$= {\left( {\dfrac{{25}}{{16}}} \right)^{ - \dfrac{3}{2}}} \\ = {\left( {\dfrac{{16}}{{25}}} \right)^{\dfrac{3}{2}}} \\$
Using exponential identity which state ${a^{\dfrac{m}{n}}} = {\left( {{a^m}} \right)^{\dfrac{1}{n}}}$ by consider m as 3 and n as 2, we get
$= {\left( {{{\left( {\dfrac{{16}}{{25}}} \right)}^3}} \right)^{\dfrac{1}{2}}}$
Power get split into numerator and denominator if a power exist for some fraction
$= {\left( {\dfrac{{{{16}^3}}}{{{{25}^3}}}} \right)^{\dfrac{1}{2}}}$
Since,${16^3} = {4^6}$and ${25^3} = {5^6}$
$= {\left( {\dfrac{{{4^6}}}{{{5^6}}}} \right)^{\dfrac{1}{2}}} \\ = \dfrac{{{4^{\dfrac{6}{2}}}}}{{{5^{\dfrac{6}{2}}}}} \\ = \dfrac{{{4^3}}}{{{5^3}}} \\ = \dfrac{{64}}{{125}} \\$
Therefore, ${\left( {\dfrac{{25}}{{16}}} \right)^{ - \dfrac{3}{2}}}$in the simplest form is equal to
$\dfrac{{64}}{{125}}$
A Rational number is a number which can be expressed in the form of $\dfrac{p}{q}$, where p and q are any integer value and q is not equal to 0. Such a number is known as a rational number.
Example: On the off chance that a number is expanded by$10{\text{ }}\%$, at that point it becomes 1.1 times of itself.