Negative Exponents: Applying the formula

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 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 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, 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:

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