Powers - special cases - Examples, Exercises and Solutions

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 zero exponent rule.

$X^0=1$We obtain

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

We use the zero exponent rule.

$X^0=1$We obtain:

$\big( \frac{7}{125}\big)^0=1$Therefore, the correct answer is option B.

Due to the fact that raising any number (except zero) to the power of zero will yield the result 1:

$X^0=1$It is thus clear that:

$(\frac{7}{4})^0=1$Therefore, the correct answer is option C.

Due to the fact that raising any number (except zero) to the power of zero will give the result 1:

$X^0=1$Let's examine the expression of the problem:

$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0$The expression inside of the parentheses is clearly not 0 (it can be calculated numerically and verified)

Therefore, the result of raising to the power of zero will give the result 1, that is:

$(300\cdot\frac{5}{3}\cdot\frac{2}{7})^0 =1$Therefore, the correct answer is option A.

We begin by applying the power rule to the products within the parentheses:

$(z\cdot t)^n=z^n\cdot t^n$That is, the power applied to a product within parentheses is applied to each of the terms when the parentheses are opened,

We apply the rule to the given problem:

$(8\cdot9\cdot5\cdot3)^{-2}=8^{-2}\cdot9^{-2}\cdot5^{-2}\cdot3^{-2}$Therefore, the correct answer is option c.

Note:

Whilst it could be understood that the above power rule applies only to two terms of the product within parentheses, in reality, it is also valid for the power over a multiplication of multiple terms within parentheses, as was seen in the above problem.

A good exercise is to demonstrate that if the previous property is valid for a power over a product of two terms within parentheses (as formulated above), then it is also valid for a power over several terms of the product within parentheses (for example - three terms, etc.).

We begin by using the negative exponent rule:

$b^{-n}=\frac{1}{b^n}$We apply it to the given expression and obtain the following:

$(3a)^{-2}=\frac{1}{(3a)^2}$We then use the power rule for parentheses:

$(x\cdot y)^n=x^n\cdot y^n$We apply it to the denominator of the expression and obtain the following:

$\frac{1}{(3a)^2}=\frac{1}{3^2a^2}=\frac{1}{9a^2}$Let's summarize the solution to the problem:

$(3a)^{-2}=\frac{1}{(3a)^2} =\frac{1}{9a^2}$

Therefore, the correct answer is option A.

We begin by using the rule for multiplying exponents. (the multiplication between terms with identical bases):

$a^m\cdot a^n=a^{m+n}$We then apply it to the problem:

$7^5\cdot7^{-6}=7^{5+(-6)}=7^{5-6}=7^{-1}$When in a first stage we begin by applying the aforementioned rule and then continue on to simplify the expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression obtained in the previous step:

$7^{-1}=\frac{1}{7^1}=\frac{1}{7}$We then summarise the solution to the problem: $7^5\cdot7^{-6}=7^{-1}=\frac{1}{7}$Therefore, the correct answer is option B.

We begin by using the power rule of exponents; for the multiplication of terms with identical bases:

$a^m\cdot a^n=a^{m+n}$We apply it to the given problem:

$12^4\cdot12^{-6}=12^{4+(-6)}=12^{4-6}=12^{-2}$When in a first stage we apply the aforementioned rule and then simplify the subsequent expression in the exponent,

Next, we use the negative exponent rule:

$a^{-n}=\frac{1}{a^n}$We apply it to the expression that we obtained in the previous step:

$12^{-2}=\frac{1}{12^2}=\frac{1}{144}$Lastly we summarise the solution to the problem: $12^4\cdot12^{-6}=12^{-2} =\frac{1}{144}$Therefore, the correct answer is option A.

First, we will focus on the exercise with a fraction in the denominator. We will solve it using the formula:

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

$\frac{x^7}{x^6}=x^{7-6}=x^1$

Therefore, we get:

$\frac{1}{x}$

We know that a product raised to the 0 is equal to 1 and therefore:

$\frac{x^0}{x^1}=x^{(0-1)}=x^{-1}$

We begin by using the power rule of negative exponents.

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

$$$4^{-1}=\frac{1}{4^1}=\frac{1}{4}$We can therefore deduce that the correct answer is option B.

We begin by using the power rule of negative exponents.

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

$$$2^{-5}=\frac{1}{2^5}=\frac{1}{32}$We can therefore deduce that the correct answer is option A.

We begin by using the power property for a negative exponent:

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

$(-7)^{-3}=\frac{1}{(-7)^3}$We then subsequently notice that each whole number inside the parentheses is raised to a negative power (that is, the number and its negative coefficient together) When using the previously mentioned power property: We are careful to take this into account,

We then continue by simplifying the expression in the denominator of the fraction, remembering the exponentiation property for the power of terms in multiplication:

$(a^m)^n=a^{m\cdot n}$We apply the resulting expression

$\frac{1}{(-7)^3}=\frac{1}{(-1\cdot7)^3}=\frac{1}{(-1)^3\cdot7^3}=\frac{1}{-1\cdot7^3}=\frac{1}{-7^3}=-\frac{1}{7^3}$

In summary we are able to deduce that the solution to the problem is as follows:

$(-7)^{-3}=\frac{1}{(-7)^3}=\frac{1}{-7^3}=-\frac{1}{7^3}$

$$$$Therefore, the correct answer is option B.

Using the rules of negative exponents: how to raise a number to a negative exponent:

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

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

We begin by using the negative exponent rule.

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

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