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

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

Answers (1)

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
\(\displaystyle{\frac{{{x}^{{{4}}}{\left({3}{x}\right)}^{{{2}}}}}{{{x}^{{{3}}}}}}={\frac{{{x}^{{{4}}}{\left({3}^{{{2}}}{x}^{{{2}}}\right)}}}{{{x}^{{{3}}}}}}\)
\(\displaystyle={\frac{{{3}^{{{2}}}{x}^{{{4}+{2}}}}}{{{x}^{{{3}}}}}}\)
\(\displaystyle={\frac{{{3}^{{{2}}}{x}^{{{6}}}}}{{{x}^{{{3}}}}}}\)
\(\displaystyle={9}{x}^{{{6}-{3}}}\)
\(\displaystyle={9}{x}^{{{3}}}\)
Answer:
The simplified form of the expression, \(\displaystyle{\frac{{{x}^{{{4}}}{\left({3}{x}\right)}^{{{2}}}}}{{{x}^{{{3}}}}}}={9}{x}^{{{3}}}\)
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