Exponents Rules - Examples, Exercises and Solutions

We use the zero exponent rule.

$X^0=1$We obtain

$112^0=1$Therefore, the correct answer is option C.

To solve the given expression $(2^3)^6$, we apply the power of a power rule $(a^m)^n = a^{m \cdot n}$. Here, $a = 2$, $m = 3$, and $n = 6$.

Thus, we calculate the exponent:

$3 \cdot 6 = 18$

So, $(2^3)^6 = 2^{18}$.

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:

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

To solve $(4^3)^2$, we use the power of a power rule which states that $(a^m)^n = a^{m \cdot n}$.

Here, $a = 4$, $m = 3$, and $n = 2$.

So, we calculate $4^{3 \cdot 2}$,

which simplifies to $4^6$.

We use the power property:

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

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

We use the formula:

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

Therefore, we obtain:

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

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.

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{3^5}{3^2} = 3^{5-2}$

Simplifying, we get $3^3$

Using the quotient rule for exponents: $\frac{a^m}{a^n} = a^{m-n}$.

Here, we have $\frac{5^6}{5^4} = 5^{6-4}$. Simplifying, we get $5^2$.

Note that in the fraction and its denominator, there are terms with the same base, so we will use the law of exponents for division between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$ Let's apply it to the problem:

$\frac{9^9}{9^3}=9^{9-3}=9^6$$$Therefore, the correct answer is b.

In the exercise of multiplying powers, we will add up all the powers of the same product, in this case the terms a, b

We use the formula:

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

We are going to focus on the term a:

$a^3\times a^2=a^{3+2}=a^5$

We are going to focus on the term b:

$b^4\times b^5=b^{4+5}=b^9$

Therefore, the exercise that will be obtained after simplification is:

$a^5\times b^9$

First, we'll enter the same fraction using the multiplication law between fractions, by multiplying numerator by numerator and denominator by denominator:

$\frac{x}{y}\cdot\frac{w}{z}=\frac{x\cdot w}{y\cdot z}$

Let's return to the problem and apply the above law:

$\frac{a^{12}}{a^9}\cdot\frac{a^3}{a^4}=\frac{a^{12}\cdot a^3}{a^{^9}\cdot a^4}$

From here on we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.

Now we'll notice that both in the numerator and denominator, multiplication is performed between terms with identical bases, therefore we'll use the power law for multiplication between terms with the same base:

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

Note that this law can only be used to calculate multiplication between terms with identical bases.

Let's return to the problem and calculate separately the results of multiplication in the numerator and denominator:

$\frac{a^{12}a^3}{a^{^9}a^4}=\frac{a^{12+3}}{a^{9+4}}=\frac{a^{15}}{a^{13}}$

where in the last step we calculated the sum of the exponents.

Now, we'll notice that we need to perform division (fraction=division operation between numerator and denominator) between terms with identical bases, therefore we'll use the power law for division between terms with the same base:

$\frac{b^m}{b^n}=b^{m-n}$

Note that this law can only be used to calculate division between terms with identical bases.

Let's return to the problem and apply the above law:

$\frac{a^{15}}{a^{13}}=a^{15-13}=a^2$

where in the last step we calculated the result of subtraction in the exponent.

We got the most simplified expression possible and therefore we're done,

therefore the correct answer is D.

Let's start with multiplying the fractions, remembering that multiplication of fractions is performed by multiplying numerator by numerator and denominator by denominator:

$\frac{b^{22}}{b^{20}}\cdot\frac{b^{30}}{b^{20}}=\frac{b^{22}\cdot b^{30}}{b^{20}\cdot b^{20}}$

Next, we'll notice that both in the numerator and denominator, multiplication occurs between terms with identical bases, so we'll use the power law for multiplying terms with identical bases:

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

We emphasize that this law can only be used when multiplication is performed between terms with identical bases.

From here on, we will no longer indicate the multiplication sign, but use the conventional writing method where placing terms next to each other means multiplication.
Let's return to the problem and apply the above power law separately to the fraction's numerator and denominator:

$\frac{b^{22}b^{30}}{b^{20}b^{20}}=\frac{b^{22+30}}{b^{20+20}}=\frac{b^{52}}{b^{40}}$

where in the final step we calculated the sum of the exponents in the numerator and denominator.

Now we notice that division is required between two terms with identical bases, so we'll use the power law for division between terms with identical bases:

$\frac{c^m}{c^n}=c^{m-n}$

We emphasize that this law can only be used when division is performed between terms with identical bases.

Let's return to the problem and apply the above power law:

$\frac{b^{52}}{b^{40}}=b^{52-40}=b^{12}$

where in the final step we calculated the subtraction between the exponents.

We have obtained the most simplified expression and therefore we are done.

Therefore, the correct answer is C.

We use the formula:

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

and therefore we obtain:

$(a^5)^7=a^{5\times7}=a^{35}$

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.