Applying Combined Exponents Rules: Presenting powers in the denominator as powers with negative exponents

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.

First, let's recall the negative exponent rule:

$b^{-n}=\frac{1}{b^n}$We'll apply it to the expression we received:

$10^{-5}=\frac{1}{10^5}=\frac{1}{100000}=0.00001$In the final steps, we performed the exponentiation in the numerator and then wrote the answer as a decimal.

Therefore, the correct answer is option A.

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.

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 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.

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.

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.

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.

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.

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.