Rules of Roots Combined: Same base and different indicator

Solve the following exercise:

$\sqrt[6]{12}\cdot\sqrt[3]{12}=$

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Let's start from converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{6}]{12}\cdot\sqrt[\textcolor{blue}{3}]{12}= \\ \downarrow\\ 12^{\frac{1}{\textcolor{red}{6}}}\cdot12^{\frac{1}{\textcolor{blue}{3}}}=$We continue, since a multiplication of two terms with identical bases is performed - we use the law of exponents shown in B:

$12^{\frac{1}{6}}\cdot12^{\frac{1}{3}}= \\ \boxed{12^{\frac{1}{6}+\frac{1}{3}}}$Therefore, the correct answer is answer C.

$12^{\frac{1}{6}+\frac{1}{3}}$

Solve the following exercise:

$\sqrt[3]{2^2}\cdot\sqrt[3]{2}=$

To simplify the given expression we use two laws of exponents:

A. The law of roots (expanded):

$\sqrt[\textcolor{blue}{n}]{a^{\textcolor{red}{m}}}=a^{\frac{\textcolor{red}{m}}{\textcolor{blue}{n}}} =(\sqrt[\textcolor{blue}{n}]{a})^{\textcolor{red}{m}}$

$$B. The law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Let's start from the root level to write exponents using the law of exponents shown in A:

$\sqrt[\textcolor{blue}{3}]{2^{\textcolor{red}{2}}}\cdot\sqrt[\textcolor{blue}{3}]{2}= \\ \sqrt[\textcolor{blue}{3}]{2^{\textcolor{red}{2}}}\cdot\sqrt[\textcolor{blue}{3}]{2^{\textcolor{red}{1}}}= \\ \downarrow\\ 2^{\frac{\textcolor{red}{2}}{\textcolor{blue}{3}}}\cdot2^{\frac{\textcolor{red}{1}}{\textcolor{blue}{3}}} =$We continue, since multiplication is performed between two terms with identical bases - we use the law of exponents shown in B:

$2^{\frac{2}{3}}\cdot2^{\frac{1}{3}}= \\ 2^{\frac{2}{3}+\frac{1}{3}}=$We continue and perform (separately) the operation of combining the numerators in the exponent fraction that was obtained, this is done by expanding each of the numerators to the common denominator - the number 3, then we perform the addition and subtraction operations in the numerator of the fraction:

$\frac{2}{3}+\frac{1}{3}=\\ \frac{2+1}{3}=\\ \frac{3}{3}=\\ 1$In other words - we get that:

$2^{\frac{2}{3}+\frac{1}{3}}=\\ 2^{1}=\\ \boxed{2}$ Let's summarize the process of simplifying the expression:

$\sqrt[3]{2^2}\cdot\sqrt[3]{2}= \\ \downarrow\\ 2^{\frac{2}{3}+\frac{1}{3}}=\\ \boxed{2}$Therefore, the correct answer is answer A.

$2$

Solve the following exercise:

$\sqrt[4]{6}\cdot\sqrt[6]{6}=$

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Let's start by converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{4}]{6}\cdot\sqrt[\textcolor{blue}{6}]{6}= \\ \downarrow\\ 6^{\frac{1}{\textcolor{red}{4}}}\cdot6^{\frac{1}{\textcolor{blue}{6}}}=$We continue, since a multiplication is performed between two terms with identical bases - we use the law of exponents shown in B:

$6^{\frac{1}{4}}\cdot6^{\frac{1}{6}}= \\ 6^{\frac{1}{4}+\frac{1}{6}}=$We continue and perform (separately) the operation of adding the exponents which are in the exponent of the expression in the simplified expression, this is done by expanding each of the exponents to the common denominator - the number 12 (which is the smallest common denominator), then we perform the addition and simplification operations in the exponent's numerator:

$\frac{1}{4}+\frac{1}{6}=\\ \frac{1\cdot3+1\cdot2}{12}=\\ \frac{3+2}{12}=\\ \frac{5}{12}\\$In other words - we get that:

$6^{\frac{1}{4}+\frac{1}{6}}=\\ \boxed{6^{\frac{5}{12}}}$To summarize the simplification process:

$\sqrt[4]{6}\cdot\sqrt[6]{6}= \\ \downarrow\\ 6^{\frac{1}{4}}\cdot6^{\frac{1}{6}}= \\ 6^{\frac{1}{4}+\frac{1}{6}}=\\ \boxed{6^{\frac{5}{12}}}$

Therefore, the correct answer is answer D.

$6^{\frac{5}{12}}$

Solve the following exercise:

$\sqrt[4]{8}\cdot\sqrt[7]{8}=$

To simplify the given expression, we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$

Let's start by converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{4}]{8}\cdot\sqrt[\textcolor{blue}{7}]{8}= \\ \downarrow\\ 8^{\frac{1}{\textcolor{red}{4}}}\cdot8^{\frac{1}{\textcolor{blue}{7}}}=$We continue, since we have a multiplication of two terms with identical bases - we use the law of exponents shown in B:

$8^{\frac{1}{4}}\cdot8^{\frac{1}{7}}= \\ \boxed{8^{\frac{1}{4}+\frac{1}{7}}}$Therefore, the correct answer is answer D.

$8^{\frac{1}{4}+\frac{1}{7}}$

Solve the following exercise:

$\sqrt[4]{3}\cdot\sqrt[6]{3}=$

In order to simplify the given expression we use two laws of exponents:

A. Defining the root as an exponent:

$\sqrt[n]{a}=a^{\frac{1}{n}}$B. The law of exponents for multiplication between factors with the same bases:

$a^m\cdot a^n=a^{m+n}$

Let's start with converting the roots to exponents using the law of exponents shown in A:

$\sqrt[\textcolor{red}{4}]{3}\cdot\sqrt[\textcolor{blue}{6}]{3}= \\ \downarrow\\ 3^{\frac{1}{\textcolor{red}{4}}}\cdot3^{\frac{1}{\textcolor{blue}{6}}}=$We continue, since multiplication is performed between two factors with the same bases - we use the law of exponents shown in B:

$3^{\frac{1}{4}}\cdot3^{\frac{1}{6}}= \\ \boxed{3^{\frac{1}{4}+\frac{1}{6}}}$Therefore, the correct answer is answer D.

$3^{\frac{1}{4}+\frac{1}{6}}$

Solve the following exercise:

$\sqrt[5]{3^3}\cdot\sqrt[5]{3^2}=$

$3$

Solve the following exercise:

$\sqrt[7]{4}\cdot\sqrt[3]{4}=$

$4^{\frac{1}{7}+\frac{1}{3}}$

Solve the following exercise:

$\sqrt[3]{5}\cdot\sqrt[3]{5}=$

$5^{\frac{2}{3}}$

Solve the following exercise:

$\sqrt{7}\cdot\sqrt{7}=$

$7$

Solve the following exercise:

$\sqrt[6]{7}\cdot\sqrt[2]{7}=$

$7^{\frac{2}{3}}$