When we have an exponent on a negative number, we can get a positive result or a negative result.

We will know this based on the exponent – whether it is even or odd.

When we have an exponent on a negative number, we can get a positive result or a negative result.

We will know this based on the exponent – whether it is even or odd.

Any number with an exponent of $0$ will be equal to $1$. (Except for $0$)

No matter which number we raise to the power of $0$, we will always get a result of 1.

In an exercise where we have a negative exponent, we turn the term into a fraction where:

the numerator will be $1$ and in the denominator, the base of the exponent with the positive exponent.

Question 1

\( \frac{1}{12^3}=\text{?} \)

Question 2

\( 19^{-2}=\text{?} \)

Question 3

\( \frac{1}{2^9}=\text{?} \)

Question 4

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

Question 5

\( \frac{1}{8^3}=\text{?} \)

$\frac{1}{12^3}=\text{?}$

First, we recall the power property for a negative exponent:

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

$\frac{1}{12^3}=12^{-3}$__Therefore, the correct answer is option A. __

$12^{-3}$

$19^{-2}=\text{?}$

To solve the exercise, we use the property of raising to a negative exponent

$a^{-n}=\frac{1}{a^n}$

We use the property to solve the exercise:

$19^{-2}=\frac{1}{19^2}$

We can continue and solve the power

$\frac{1}{19^2}=\frac{1}{361}$

$\frac{1}{361}$

$\frac{1}{2^9}=\text{?}$

We use the power property for a negative exponent:

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

$\frac{1}{2^9}=2^{-9}$

__Therefore, the correct answer is option A. __

$2^{-9}$

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

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.__

$4$

$\frac{1}{8^3}=\text{?}$

We use the power property for a negative exponent:

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

$\frac{1}{8^3}=8^{-3}$When we use this previously mentioned property in the opposite sense.

__Therefore, the correct answer is option A. __

$8^{-3}$

Question 1

\( 5^0= \)

Question 2

\( 5^{-2} \)

Question 3

\( 7^{-24}=\text{?} \)

Question 4

\( 4^{-1}=\text{?} \)

Question 5

\( 112^0=\text{?} \)

$5^0=$

We use the power property:

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

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

$1$

$5^{-2}$

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.__

$\frac{1}{25}$

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

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.__

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

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

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. __

$\frac{1}{4}$

$112^0=\text{?}$

We use the zero exponent rule.

$X^0=1$We obtain

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

1

Question 1

\( \frac{1}{(-2)^7}=? \)

Question 2

\( 5^4\cdot(\frac{1}{5})^4=\text{?} \)

Question 3

\( (\frac{7}{4})^?=1 \)

Question 4

\( \frac{9\cdot3}{8^0}=\text{?} \)

Question 5

\( (-7)^{-3}=\text{?} \)

$\frac{1}{(-2)^7}=?$

**First**, we deal with the expression in the denominator of the fraction and remember according to the property of raising an exponent to another exponent:

$(a^m)^n=a^{m\cdot n}$We obtain that:

$(-2)^7=(-1\cdot2)^7=(-1)^7\cdot2^7=-1\cdot2^7=-2^7$**We return to the problem **and apply what was said before:

$\frac{1}{(-2)^7}=\frac{1}{-2^7}=\frac{1}{-1}\cdot\frac{1}{2^7}=-\frac{1}{2^7}$When in the last step we remember that:

$\frac{1}{-1}=-1$Next, **we remember** the property of raising to a negative power

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

$-\frac{1}{2^7}=-2^{-7}$**Let's summarize** the steps of the solution:

$\frac{1}{(-2)^7}=-\frac{1}{2^7} = -2^{-7}$

__Therefore, the correct answer is option C. __

$(-2)^{-7}$

$5^4\cdot(\frac{1}{5})^4=\text{?}$

**This problem can be solved using the properties of powers for a negative power, power over a power, and the property of powers for the product between terms with identical bases, which is the natural way of solving,**

**But here we prefer to solve it in another way that is a bit faster:**

To this end, the power by power law is applied to the parentheses in which the terms are multiplied, **but in the opposite direction**:

$x^n\cdot y^n=(x\cdot y)^n$Since in the expression in the problem **there is a multiplication between two terms with identical powers**, this law can be used in its opposite sense, so we will apply this property to the problem:

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4$**Since the multiplication in the given problem is between terms with the same power, we could apply this law in the opposite direction** and write the expression as the multiplication of the bases of the terms in parentheses to which the same power is applied.

We will continue and simplify the expression in parentheses, we will do it quickly if we notice that in parentheses **there is a multiplication between two opposite numbers, then their product will give the result: 1,** we will apply this understanding to the expression we arrived at in the last step:

$\big(5\cdot\frac{1}{5}\big)^4 = 1^4=1$When in the first step we apply the previous understanding, and then use the fact that **raising the number 1 to any power will always give the result: 1**, which means that:

$1^x=1$**Summarizing** the steps to solve the problem, we get that:

$5^4\cdot(\frac{1}{5})^4=\big(5\cdot\frac{1}{5}\big)^4 =1$__Therefore, the correct answer is option b.__

1

$(\frac{7}{4})^?=1$

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

$X^0=1$Therefore, it is clear that:

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

0

$\frac{9\cdot3}{8^0}=\text{?}$

We use the formula:

$a^0=1$

$\frac{9\times3}{8^0}=\frac{9\times3}{1}=9\times3$

We know that:

$9=3^2$

Therefore, we obtain:

$3^2\times3=3^2\times3^1$

We use the formula:

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

$3^2\times3^1=3^{2+1}=3^3$

$3^3$

$(-7)^{-3}=\text{?}$

We use the power property for a negative exponent:

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

$(-7)^{-3}=\frac{1}{(-7)^3}$When we notice that ** each** whole number in parentheses is raised to a negative power (that is, the number and its negative coefficient together), by using the previously mentioned power property

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

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

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

**Summarizing** the solution to the problem, we obtained that:

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

__Therefore, the correct answer is option B. __

$-\frac{1}{7^{3}}$