Rules of Roots combined - Examples, Exercises and Solutions

$(3\times4\times5)^4=$

We use the power law for multiplication within parentheses:

$(x\cdot y)^n=x^n\cdot y^n$We apply it to the problem:

$(3\cdot4\cdot5)^4=3^4\cdot4^4\cdot5^4$Therefore, the correct answer is option b.

Note:

From the formula of the power property mentioned above, we understand that it refers not only to two terms of the multiplication within parentheses, but also for multiple terms within parentheses.

$3^44^45^4$

$(4\times7\times3)^2=$

We use the power law for multiplication within parentheses:

$$$(x\cdot y)^n=x^n\cdot y^n$We apply it to the problem:

$(4\cdot7\cdot3)^2=4^2\cdot7^2\cdot3^2$Therefore, the correct answer is option a.

Note:

From the formula of the power property mentioned above, we understand that we can apply it not only to the multiplication of two terms within parentheses, but is also for multiple terms within parentheses.

$4^2\times7^2\times3^2$

$5^4\times25=$

To solve this exercise, first we note that 25 is the result of a power and we reduce it to a common base of 5.

$\sqrt{25}=5$$25=5^2$Now, we go back to the initial exercise and solve by adding the powers according to the formula:

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

$5^4\times25=5^4\times5^2=5^{4+2}=5^6$

$5^6$

$(4^2)^3+(g^3)^4=$

We use the formula:

$(a^m)^n=a^{m\times n}$

$(4^2)^3+(g^3)^4=4^{2\times3}+g^{3\times4}=4^6+g^{12}$

$4^6+g^{12}$

$\frac{2^4}{2^3}=$

Let's keep in mind that the numerator and denominator of the fraction have terms with the same base, therefore we use the property of powers to divide between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$We apply it in the problem:

$\frac{2^4}{2^3}=2^{4-3}=2^1$Remember that any number raised to the 1st power is equal to the number itself, meaning that:

$b^1=b$Therefore, in the problem we obtain:

$2^1=2$Therefore, the correct answer is option a.

$2$

$\frac{81}{3^2}=$

First, we recognize that 81 is a power of the number 3, which means that:

$3^4=81$We replace in the problem:

$\frac{81}{3^2}=\frac{3^4}{3^2}$Keep in mind that the numerator and denominator of the fraction have terms with the same base, therefore we use the property of powers to divide between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$We apply it in the problem:

$\frac{3^4}{3^2}=3^{4-2}=3^2$$$Therefore, the correct answer is option b.

$3^2$$$

$(3^5)^4=$

To solve the exercise we use the power property:$(a^n)^m=a^{n\cdot m}$

We use the property with our exercise and solve:

$(3^5)^4=3^{5\times4}=3^{20}$

$3^{20}$

$(6^2)^{13}=$

We use the formula:

$(a^n)^m=a^{n\times m}$

Therefore, we obtain:

$6^{2\times13}=6^{26}$

$6^{26}$

$(\frac{2}{6})^3=$

We use the formula:

$(\frac{a}{b})^n=\frac{a^n}{b^n}$

$(\frac{2}{6})^3=(\frac{2}{2\times3})^3$

We simplify:

$(\frac{1}{3})^3=\frac{1^3}{3^3}$

$\frac{1\times1\times1}{3\times3\times3}=\frac{1}{27}$

$\frac{1}{27}$

$5^0=$

We use the power property:

$X^0=1$We apply it to the problem:

$5^0=1$Therefore, the correct answer is C.

$1$

$(\frac{4^2}{7^4})^2=$

$(\frac{4^2}{7^4})^2=\frac{4^{2\times2}}{7^{4\times2}}=\frac{4^4}{7^8}$

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

$(\frac{1}{4})^{-1}$

We use the power property for a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will write the fraction in parentheses as a negative power with the help of the previously mentioned power:

$\frac{1}{4}=\frac{1}{4^1}=4^{-1}$We return to the problem, where we obtained:

$\big(\frac{1}{4}\big)^{-1}=(4^{-1})^{-1}$We continue and use the power property of an exponent raised to another exponent:

$(a^m)^n=a^{m\cdot n}$And we apply it in the problem:

$(4^{-1})^{-1}=4^{-1\cdot-1}=4^1=4$Therefore, the correct answer is option d.

$4$

$5^{-2}$

We use the property of powers of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the problem:

$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$

Therefore, the correct answer is option d.

$\frac{1}{25}$

$4^{-1}=\text{?}$

We use the property of raising to a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it to the problem:

$$$4^{-1}=\frac{1}{4^1}=\frac{1}{4}$Therefore, the correct answer is option B.

$\frac{1}{4}$

$7^{-24}=\text{?}$

We use the property of raising to a negative exponent:

$a^{-n}=\frac{1}{a^n}$We apply it in the problem:

$$$$$7^{-24}=\frac{1}{7^{24}}$Therefore, the correct answer is option D.

$\frac{1}{7^{24}}$