Question

How do you simplify ${{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}$

Answer 1

Now to simplify the equation we will first use the power rule of indices. Now we will again simplify the obtained expression by converting negative powers into positive by negative exponent rule. Now we will convert the expression in radical form. Now we will factor the numbers in root and simplify the expression.

Complete step-by-step solution:
Now to simplify the equation we will have to use some properties of indices.
Let us first understand the concept of indices.
Now consider the number ${{3}^{4}}$ here we say that 3 is the base and 4 is the power to which it is raised. But what does ${{3}^{4}}$ denote? The number ${{3}^{4}}$ means multiplication of 3, 4 times.
Hence ${{3}^{4}}=3\times 3\times 3\times 3$ .
Hence in general if we have ${{a}^{m}}$ then it means $a\times a\times a...\text{n times}$.
Now let us learn some basic laws of indices.
The multiplication law of indices states that ${{a}^{m}}.{{a}^{n}}={{a}^{m+n}}$
Similarly the division law of indices states that $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$
And the power law of indices states ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ .
Now according to law of negative integer we have ${{a}^{-m}}=\dfrac{1}{{{a}^{m}}}$
Now let us understand what it means when there is a fraction in power.
Hence consider ${{a}^{\dfrac{m}{n}}}$ then the number can be written in radical form by, ${{a}^{\dfrac{m}{n}}}=\sqrt[n]{{{a}^{m}}}$
Now consider the given expression ${{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}$
Now using power law we have,
$\begin{aligned} & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( {{16}^{\dfrac{5\times \left( -3 \right)}{9}}}{{.5}^{\dfrac{7\times \left( -3 \right)}{9}}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( {{16}^{\dfrac{-5}{3}}}{{.5}^{\dfrac{-7}{3}}} \right) \\\ \end{aligned}$
Now using the law of negative exponents we have,
$\Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{{{16}^{\dfrac{5}{3}}}}.\dfrac{1}{{{5}^{\dfrac{7}{3}}}} \right)$
Now converting the indices into radicals we get,

& \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{\sqrt[3]{{{16}^{5}}}}.\dfrac{1}{\sqrt[3]{{{5}^{7}}}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{\sqrt[3]{16\times 16\times 16\times 16\times 16}}.\dfrac{1}{\sqrt[3]{5\times 5\times 5\times 5\times 5\times 5\times 5}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{16\sqrt[3]{16\times 16}}.\dfrac{1}{{{5}^{2}}\sqrt[3]{5}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{400\sqrt[3]{8\times 8\times 2\times 2\times 5}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{400\times \left( 2\times 2 \right)\sqrt[3]{20}} \right) \\\ & \Rightarrow {{\left( {{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}} \right)}^{-3}}=\left( \dfrac{1}{1600\sqrt[3]{20}} \right) \\\ \end{aligned}$$ Hence the given expression is simplified. **Note:** Note that to use multiplication and division law the base number must be the same. Hence do not make a mistake of writing ${{16}^{\dfrac{5}{9}}}{{.5}^{\dfrac{7}{9}}}$ as ${{\left( 16\times 5 \right)}^{\dfrac{5}{9}+\dfrac{7}{9}}}$ . Also note that the negative exponent law is the same as division law. Since $\dfrac{1}{{{x}^{m}}}=\dfrac{{{x}^{0}}}{{{x}^{m}}}={{x}^{0-m}}={{x}^{-m}}$.