what is the following product? assume x≥0. (sqrt3{x^{2}}cdotsqrt4{x^{3}}) \n(xsqrt{x}) \n(sqrt12{x^{5}})…

what is the following product? assume x≥0. (sqrt3{x^{2}}cdotsqrt4{x^{3}}) \n(xsqrt{x}) \n(sqrt12{x^{5}}) \n(xleft(sqrt12{x^{5}}\right)) \n(x^{6})

what is the following product? assume x≥0. (sqrt3{x^{2}}cdotsqrt4{x^{3}}) \n(xsqrt{x}) \n(sqrt12{x^{5}}) \n(xleft(sqrt12{x^{5}}\right)) \n(x^{6})

Answer

Explanation:

Step1: Rewrite radicals as exponents

Recall that $\sqrt[n]{a}=a^{\frac{1}{n}}$. So, $\sqrt[3]{x^{2}}=x^{\frac{2}{3}}$ and $\sqrt[4]{x^{3}}=x^{\frac{3}{4}}$.

Step2: Use exponent - multiplication rule

When multiplying two terms with the same base $a^m\cdot a^n=a^{m + n}$. Here, $x^{\frac{2}{3}}\cdot x^{\frac{3}{4}}=x^{\frac{2}{3}+\frac{3}{4}}$.

Step3: Find a common denominator

The common denominator of 3 and 4 is 12. So, $\frac{2}{3}+\frac{3}{4}=\frac{2\times4}{3\times4}+\frac{3\times3}{4\times3}=\frac{8}{12}+\frac{9}{12}=\frac{8 + 9}{12}=\frac{17}{12}$.

Step4: Rewrite the result as a radical

$x^{\frac{17}{12}}=x^{1+\frac{5}{12}}=x\cdot x^{\frac{5}{12}}=x\sqrt[12]{x^{5}}$

Answer:

$x\left(\sqrt[12]{x^{5}}\right)$