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

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

Procedure used:
The definition of a Rational Exponent is,
For any rational exponent m/n in lowest, where m and n are integers and $n>0$, we define

If n is even, then we require that $a\ge 0$.

This implies that $b={a}^{\frac{1}{2}}$ and therefore
$\sqrt{a}={a}^{\frac{1}{2}}$
According to the Zero and Negative Exponents rule, we can see that if $a\ne q0$ is a real number and n is a positive integer, then

From, the law of exponents, we know that
${\left(ab\right)}^{m}={a}^{m}{b}^{m}$.....(1)
${a}^{m}{a}^{n}={a}^{m+n}$.....(2)
$\frac{{a}^{m}}{{a}^{n}}={a}^{m-n}$.....(3)
Calculation:
Using the Law of Exponents (1), (2) and (3), we get that
$\frac{{x}^{4}{\left(3x\right)}^{2}}{{x}^{3}}=\frac{{x}^{4}\left({3}^{2}{x}^{2}\right)}{{x}^{3}}$
$=\frac{{3}^{2}{x}^{4+2}}{{x}^{3}}$
$=\frac{{3}^{2}{x}^{6}}{{x}^{3}}$
$=9{x}^{6-3}$
$=9{x}^{3}$
The simplified form of the expression, $\frac{{x}^{4}{\left(3x\right)}^{2}}{{x}^{3}}=9{x}^{3}$

Jeffrey Jordon