Negative Exponents: Presenting powers in the denominator as powers with negative exponents

To begin with, we must remind ourselves of the Negative Exponent rule:

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

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

To solve the given problem, we need to express $\frac{1}{5^2}$ using negative exponents. We'll apply the formula for negative exponents, which is $\frac{1}{a^n} = a^{-n}$:

• Identify the base and power in the denominator. Here, the base is $5$ and the power is $2$.
• Apply the inverse formula: $\frac{1}{5^2} = 5^{-2}$.

Thus, the equivalent expression for $\frac{1}{5^2}$ using a negative exponent is $5^{-2}$.

To solve the problem of expressing $\frac{1}{4^2}$ using powers with negative exponents:

• Identify the base in the denominator: 4 raised to the power 2.
• Apply the rule for negative exponents that states $\frac{1}{a^n} = a^{-n}$.
• Express $\frac{1}{4^2}$ as $4^{-2}$.

Thus, the expression $\frac{1}{4^2}$ can be rewritten as $4^{-2}$.

To solve this problem, we'll use the rule of negative exponents:

• Step 1: Identify that the given expression is $\frac{1}{3^2}$.
• Step 2: Recognize that $\frac{1}{3^2}$ can be rewritten using the negative exponent rule.
• Step 3: Apply the formula $\frac{1}{a^n} = a^{-n}$ to the expression $\frac{1}{3^2}$.

Now, let's work through these steps:

Step 1: We have $\frac{1}{3^2}$ where 3 is the base and 2 is the exponent.

Step 2: Using the formula, convert the denominator $3^2$ to $3^{-2}$.

Step 3: Thus, $\frac{1}{3^2} = 3^{-2}$.

Therefore, the solution to the problem is $3^{-2}$.

To solve this problem, we will rewrite the expression $\frac{1}{6^7}$ using the rules of exponents:

Step 1: Identify the given fraction.

We start with $\frac{1}{6^7}$, where the base in the denominator is 6, and the exponent is 7.

Step 2: Apply the formula for negative exponents.

The formula $a^{-n} = \frac{1}{a^n}$ allows us to rewrite a reciprocal power as a negative exponent. This means the expression $\frac{1}{6^7}$ can be rewritten as $6^{-7}$.

Step 3: Conclude with the answer.

By transforming $\frac{1}{6^7}$ to its equivalent form using negative exponents, the expression becomes $6^{-7}$.

Therefore, the correct expression is $\boxed{6^{-7}}$, which corresponds to choice 2 in the given options.

To solve this problem, we will use the properties of exponents. Specifically, we will convert the expression $\frac{1}{20^2}$ into a form that uses a negative exponent. The general relationship is that $\frac{1}{a^n} = a^{-n}$.

Applying this rule to the given expression:

• Step 1: Identify the current form, which is $\frac{1}{20^2}$.
• Step 2: Apply the negative exponent rule: $\frac{1}{20^2} = 20^{-2}$.
• Step 3: This expression, $20^{-2}$, represents $\frac{1}{20^2}$ using a negative exponent.

Therefore, the expression $\frac{1}{20^2}$ can be expressed as $20^{-2}$, which aligns with choice 1.

We use the negative exponent rule.

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

We apply it to the problem in the opposite sense.:

$$$$$\frac{1}{8^3}=8^{-3}$

Therefore, the correct answer is option A.

We use the power property for a negative exponent:

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

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

Therefore, the correct answer is option A.

First, let's note that 27 is a power of the number 3:

$27=3^3$Using this fact gives us a situation where in the fraction's numerator and denominator we get terms with identical bases, let's apply this to the problem:

$\frac{27}{3^8}=\frac{3^3}{3^8}$Now let's recall the law of exponents for division between terms without identical bases:

$\frac{a^m}{a^n}=a^{m-n}$ Let's apply this law to the last expression we got:

$\frac{3^3}{3^8}=3^{3-8}=3^{-5}$where in the first stage we applied the above law and in the second stage we simplified the expression we got in the exponent,

Let's summarize the solution steps, we got:

$\frac{27}{3^8}=\frac{3^3}{3^8}=3^{-5}$Therefore the correct answer is answer D.

To begin with we deal with the expression in the denominator of the fraction. Making note of the power rule for exponents (raising an exponent to another exponent):

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

$(-2)^7=(-1\cdot2)^7=(-1)^7\cdot2^7=-1\cdot2^7=-2^7$

We then return to the initial problem and apply the above information:

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

In the last step we remember that:

$\frac{1}{-1}=-1$

Next, we remember the Negative Exponent rule ( raising exponents 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.

This question is actually a proof of the law of exponents for negative exponents, we will prove it simply using two other laws of exponents:

a. The zero exponent law, which states that raising any number to the power of 0 (except 0) will give the result 1:

$X^0=1$

b. The law of exponents for division between terms with identical bases:

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

Let's return to the problem and pay attention to two things, the first is that in the denominator of the fraction there is a term with base $a$and the second thing is that according to the zero exponent law mentioned above in a' we can always write the number 1 as any number (except 0) to the power of 0, particularly in this problem, given that $a\neq0$ we can claim that:

$1=a^0$

Let's apply this to the problem:

$\frac{1}{a^n}=\frac{a^0}{a^n}$

Now that we have in the numerator and denominator of the fraction terms with identical bases, we can use the law of division between terms with identical bases mentioned in b' in the problem:

$\frac{a^0}{a^n}=a^{0-n}=a^{-n}$

Let's summarize the steps above, we got that:

$\frac{1}{a^n}=\frac{a^0}{a^n}=a^{-n}$

In other words, we proved the law of exponents for negative exponents and understood why the correct answer is answer c.

First, let's note that 4 is a power of 2:

$4=2^2$therefore we can perform a conversion to a common base for all terms in the problem,

Let's apply this:

$\frac{2}{4^{-2}}=\frac{2}{(2^2)^{-2}}$Next, we'll use the power law for power of power:

$(a^m)^n=a^{m\cdot n}$and we'll apply this law to the denominator term we got in the last step:

$\frac{2}{(2^2)^{-2}}=\frac{2}{2^{2\cdot(-2)}}=\frac{2}{2^{-4}}$where in the first step we applied the above law to the denominator and in the second step we simplified the expression we got,

Next, we'll use the power law for division between terms with identical bases:

$\frac{a^m}{a^n}=a^{m-n}$and we'll apply this law to the last expression we got:

$\frac{2}{2^{-4}}=2^{1-(-4)}=2^{1+4}=2^5$

Therefore the correct answer is answer B.

First, let's note that:

a.

$10=5\cdot2$

For this, we'll recall the law of exponents for multiplication in parentheses:

$(a\cdot b)^n=a^n\cdot b^n$

According to this, we get that:

$(-5)^3=(-1\cdot5)^3=(-1)^3\cdot5^3=-1\cdot5^3=-5^3$

We want to use the understanding in 'a' to get terms with identical bases in the numerator and denominator,

Let's return to the problem and apply the understandings from 'a' and 'b':

$\frac{10}{(-5)^3}=\frac{2\cdot5}{-5^3}=\frac{2}{-1}\cdot\frac{5}{5^3}=-2\cdot\frac{5}{5^3}$

Where in the first stage we used 'a' in the numerator and 'b' in the fraction's denominator, in the next stage we presented the fraction as a multiplication of fractions according to the rule for multiplying fractions, then we simplified the first fraction in the multiplication.

Now we'll use the law of exponents for division between terms with identical bases:

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

Let's apply this law to the expression we got:

$-2\cdot\frac{5}{5^3}=-2\cdot5^{1-3}=-2\cdot5^{-2}$

where in the first stage we applied this law to the fraction in the multiplication and then simplified the expression we got,

Let's summarize the solution steps:

$\frac{10}{(-5)^3} =-2\cdot\frac{5}{5^3} =-2\cdot5^{-2}$

Therefore, the correct answer is answer b.