# Radicals and Exponents Simplify the expression. frac{x^{4}(3x)^{2}}{x^{3}}

Question
Radicals and Exponents Simplify the expression.
$$\displaystyle{\frac{{{x}^{{{4}}}{\left({3}{x}\right)}^{{{2}}}}}{{{x}^{{{3}}}}}}$$

2020-10-19
Procedure used:
The definition of a Rational Exponent is,
For any rational exponent m/n in lowest, where m and n are integers and $$\displaystyle{n}{>}{0}$$, we define
$$\displaystyle{a}^{{\frac{{m}}{{n}}}}={\left(\sqrt{{{n}}}{\left\lbrace{a}\right\rbrace}\right)}^{{{m}}}{]}{\quad\text{or}\quad}\equiv{a}\le{n}{t}{l}{y}{\left[{a}^{{\frac{{m}}{{n}}}}=\sqrt{{{n}}}{\left\lbrace{a}\right\rbrace}^{{{m}}}.\right.}$$
If n is even, then we require that $$\displaystyle{a}\geq{0}$$.
$$\displaystyle\sqrt{{{a}}}={b}\ \text{means}\ {b}^{{{2}}}={a}\ \text{and}\ {b}\geq\ {0}$$
This implies that $$\displaystyle{b}={a}^{{\frac{{1}}{{2}}}}$$ and therefore
$$\displaystyle\sqrt{{{a}}}={a}^{{\frac{{1}}{{2}}}}$$
According to the Zero and Negative Exponents rule, we can see that if $$\displaystyle{a}\ne{q}{0}$$ is a real number and n is a positive integer, then
$$\displaystyle{a}^{{{0}}}={1}\ \text{and}\ {a}^{{-{n}}}={\frac{{{1}}}{{{a}^{{{n}}}}}}$$
From, the law of exponents, we know that
$$\displaystyle{\left({a}{b}\right)}^{{{m}}}={a}^{{{m}}}{b}^{{{m}}}$$.....(1)
$$\displaystyle{a}^{{{m}}}{a}^{{{n}}}={a}^{{{m}+{n}}}$$.....(2)
$$\displaystyle{\frac{{{a}^{{{m}}}}}{{{a}^{{{n}}}}}}={a}^{{{m}-{n}}}$$.....(3)
Calculation:
Using the Law of Exponents (1), (2) and (3), we get that
$$\displaystyle{\frac{{{x}^{{{4}}}{\left({3}{x}\right)}^{{{2}}}}}{{{x}^{{{3}}}}}}={\frac{{{x}^{{{4}}}{\left({3}^{{{2}}}{x}^{{{2}}}\right)}}}{{{x}^{{{3}}}}}}$$
$$\displaystyle={\frac{{{3}^{{{2}}}{x}^{{{4}+{2}}}}}{{{x}^{{{3}}}}}}$$
$$\displaystyle={\frac{{{3}^{{{2}}}{x}^{{{6}}}}}{{{x}^{{{3}}}}}}$$
$$\displaystyle={9}{x}^{{{6}-{3}}}$$
$$\displaystyle={9}{x}^{{{3}}}$$
The simplified form of the expression, $$\displaystyle{\frac{{{x}^{{{4}}}{\left({3}{x}\right)}^{{{2}}}}}{{{x}^{{{3}}}}}}={9}{x}^{{{3}}}$$

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