Negative Exponents: Applying the formula

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.

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.

We use the power property of a negative exponent:

$a^{-n}=\frac{1}{a^n}$We will rewrite the fraction in parentheses as a negative power:

$\frac{1}{7}=7^{-1}$Let's return to the problem, where we had:

$\bigg( \big( \frac{1}{7}\big)^{-1}\bigg)^4=\big((7^{-1})^{-1} \big)^4$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:

$\big((7^{-1})^{-1} \big)^4 =(7^{-1\cdot-1})^4=(7^1)^4=7^{1\cdot4}=7^4$Therefore, the correct answer is option c

We use the law of exponents to multiply terms with identical bases:

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

$7^x\cdot7^{-x}=7^{x+(-x)}=7^{x-x}=7^0$In the first stage we apply the above power rule and in the following stages we simplify the expression obtained in the exponent,

Subsequently, we use the zero power rule:

$X^0=1$We obtain:

$7^0=1$Lastly we summarize the solution to the problem.

$7^x\cdot7^{-x}=7^{x-x}=7^0 =1$Therefore, the correct answer is option B.

First, let's note that the first term in the multiplication in the problem has an exponent of 0, and any number (different from zero) raised to the power of zero equals 1, meaning:

$X^0=1$Therefore, we get that the expression in the problem is:

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= 1\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}=(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

Later, we will use the law of exponents for negative exponents:

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

And before we proceed to solve the problem let's understand this law in a slightly different, indirect way:

Let's note that if we treat this law as an equation (which it indeed is in every way), and multiply both sides of the equation by the common denominator which is:

$a^n$we get:

$a^{-n}=\frac{1}{a^n}\\ \frac{a^n}{1} =\frac{1}{a^n}\hspace{8pt} \text{/}\cdot a^n\\ a^n\cdot a^{-n}=1$

Where in the first stage we remembered that any number can be represented as itself divided by 1, we applied this to the left side of the equation, then we multiplied by the common denominator and to know by how much we multiplied each numerator (after reduction with the common denominator) we addressed the question "by how much did we multiply the current denominator to get the common denominator?".

Let's look at the result we got:

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

Meaning that $a^n,\hspace{4pt}a^{-n}$are reciprocal numbers, or in other words:

$a^n$ is reciprocal to $a^{-n}$ (and vice versa),

and specifically:

$a,\hspace{4pt}a^{-1}$ are reciprocal to each other,

We can apply this understanding to the problem if we also remember that the reciprocal of a fraction is obtained by switching the numerator and denominator, meaning that the fractions:

$\frac{a}{b},\hspace{4pt}\frac{b}{a}$ are reciprocal fractions - which can be easily understood, since their multiplication will clearly give the result 1,

And if we combine this with the previous understanding, we can easily conclude that:

$\big(\frac{a}{b}\big)^{-1}=\frac{b}{a}$

In other words, raising a fraction to the power of negative one will give a result that is the reciprocal fraction, obtained by switching the numerator and denominator.

Let's return to the problem, the expression we got in the last stage is:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}$

We'll use what was explained earlier and note that the fraction in parentheses in the second term of the multiplication is the reciprocal fraction to the fraction in parentheses in the first term of the multiplication, meaning that:$\frac{13}{2}= \big(\frac{2}{13} \big)^{-1}$Therefore we can simply calculate the expression we got in the last stage by converting to a common base using the above understanding:

$(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5}$

Where we actually just replaced the fraction in parentheses in the second term of the multiplication with its reciprocal raised to the power of negative one as mentioned earlier,

Next we'll recall the law of exponents for power of a power:

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

And we'll apply this law to the expression we got in the last stage:

$(\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{(-1)\cdot(-5)}=(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

Where in the first stage we applied the above law of exponents and then simplified the expression that resulted,

Let's summarize the solution of the problem so far, we got that:

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} = (\frac{2}{13})^{-2}\cdot\big( (\frac{2}{13})^{-1}\big)^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}$

In the next stage we'll recall the law of exponents for multiplication of terms with identical bases:

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

We'll apply this law to the expression we got in the last stage:

$(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5} =(\frac{2}{13}\big)^{-2+5}=(\frac{2}{13}\big)^{3}$

Where in the first stage we applied the above law of exponents and then simplified the expression,

Let's summarize the solution of the problem so far, we got that:

$(\frac{13}{2})^0\cdot(\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5}= (\frac{2}{13})^{-2}\cdot(\frac{13}{2})^{-5} =(\frac{2}{13})^{-2}\cdot (\frac{2}{13})^{5}=(\frac{2}{13}\big)^{3}$

Therefore the correct answer is answer A.

To solve the problem $300^{-4} \cdot \left(\frac{1}{300}\right)^{-4}$, let's follow these steps:

• Step 1: Simplify the reciprocal power expression.
  The expression $\left(\frac{1}{300}\right)^{-4}$ can be simplified using the rule for reciprocals, which states $\left(\frac{1}{a}\right)^{-n} = a^n$. Thus, $\left(\frac{1}{300}\right)^{-4} = 300^4$.
• Step 2: Combine the powers.
  Now the expression becomes $300^{-4} \cdot 300^4$. Using the rule $a^m \cdot a^n = a^{m+n}$, we have $300^{-4+4} = 300^0$.
• Step 3: Calculate the final result.
  By the identity $a^0 = 1$, any non-zero number raised to the power of zero equals 1. Therefore, $300^0 = 1$.

Thus, the solution to the problem is $\boxed{1}$.

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:

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