Evaluating Exponential And Zero Exponent Expressions

In this comprehensive guide, we will delve into the evaluation of various mathematical expressions involving exponents, negative exponents, and zero exponents. Mastering these concepts is crucial for a solid foundation in algebra and higher-level mathematics. We will systematically break down each expression, providing clear explanations and step-by-step solutions. From simple fractional exponents to more complex combinations of operations, this exploration will equip you with the tools and understanding necessary to confidently tackle a wide range of mathematical problems.

a. Evaluating ${\left(\frac{-7}{10}\right)^3}$

To evaluate this expression, we need to understand the meaning of an exponent. An exponent indicates how many times the base is multiplied by itself. In this case, the base is ${\frac{-7}{10}}$ and the exponent is 3, which means we multiply ${\frac{-7}{10}}$ by itself three times:

${\left(\frac{-7}{10}\right)^3 = \frac{-7}{10} \times \frac{-7}{10} \times \frac{-7}{10}}$

When multiplying fractions, we multiply the numerators together and the denominators together:

${= \frac{(-7) \times (-7) \times (-7)}{10 \times 10 \times 10}}$

${= \frac{-343}{1000}}$

Therefore, ${\left(\frac{-7}{10}\right)^3 = \frac{-343}{1000}}$. Understanding the properties of exponents is crucial in simplifying and evaluating such expressions. When a negative fraction is raised to an odd power, the result is negative. This is because a negative number multiplied by a negative number results in a positive number, but multiplying by another negative number makes the result negative again. On the other hand, if the exponent were an even number, the result would be positive. This is a fundamental concept in dealing with exponents and negative numbers. To further illustrate, consider ${\left(\frac{-1}{2}\right)^3}$, which equals ${\frac{-1}{8}}$, or ${\left(\frac{-1}{2}\right)^2}$, which equals ${\frac{1}{4}}$. This distinction between odd and even exponents is crucial for accurately evaluating expressions.

d. Evaluating ${65(47^0 + 33^0)}$

This expression involves the concept of a zero exponent. Any non-zero number raised to the power of 0 is equal to 1. This is a fundamental rule in exponents. Therefore, ${47^0 = 1}$ and ${33^0 = 1}$. Substituting these values into the expression, we get:

${65(47^0 + 33^0) = 65(1 + 1)}$

${= 65(2)}$

${= 130}$

Thus, ${65(47^0 + 33^0) = 130}$. The rule that any non-zero number raised to the power of zero equals one is a cornerstone of exponential operations. It might seem counterintuitive at first, but it is a necessary rule to maintain consistency within the mathematical system of exponents. For instance, consider the general rule ${a^m / a^n = a^{m-n}}$. If we let ${m = n}$, we get ${a^m / a^m = a^{m-m} = a^0}$. Since any number divided by itself is 1, ${a^0}$ must equal 1 to make the rule consistent. This principle is essential for simplifying expressions and solving equations involving exponents.

g. Evaluating ${\left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2}}$

This expression involves negative exponents and division. A negative exponent indicates the reciprocal of the base raised to the positive exponent. That is, ${a^{-n} = \frac{1}{a^n}}$. Applying this rule, we can rewrite the expression as:

${\left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2} = \frac{1}{\left(\frac{15}{47}\right)^4} \div \frac{1}{\left(\frac{15}{47}\right)^2}}$

Dividing by a fraction is the same as multiplying by its reciprocal. Therefore, we can rewrite the division as multiplication:

${= \frac{1}{\left(\frac{15}{47}\right)^4} \times \left(\frac{15}{47}\right)^2}$

${= \frac{\left(\frac{15}{47}\right)^2}{\left(\frac{15}{47}\right)^4}}$

Now, we can use the rule for dividing exponential expressions with the same base: ${\frac{a^m}{a^n} = a^{m-n}}$. In this case:

${= \left(\frac{15}{47}\right)^{2-4}}$

${= \left(\frac{15}{47}\right)^{-2}}$

Applying the negative exponent rule again:

${= \left(\frac{47}{15}\right)^{2}}$

${= \frac{47^2}{15^2}}$

${= \frac{2209}{225}}$

Thus, ${\left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2} = \frac{2209}{225}}$. This evaluation demonstrates the practical application of negative exponents and the rules of exponent division. A negative exponent essentially indicates the reciprocal of the base raised to the corresponding positive exponent. The rule ${a^{-n} = \frac{1}{a^n}}$ is fundamental in simplifying such expressions. Moreover, when dividing exponential expressions with the same base, we subtract the exponents, as shown by the rule ${\frac{a^m}{a^n} = a^{m-n}}$. These rules allow us to efficiently manipulate and simplify complex expressions involving exponents.

j. Evaluating ${\left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4}$

To evaluate this expression, we first calculate each exponential term separately:

${\left(\frac{9}{16}\right)^2 = \frac{9^2}{16^2} = \frac{81}{256}}$

${\left(\frac{2}{3}\right)^4 = \frac{2^4}{3^4} = \frac{16}{81}}$

Now, we multiply the results:

${\frac{81}{256} \times \frac{16}{81}}$

We can simplify by canceling out the common factor of 81:

${= \frac{1}{256} \times 16}$

Now, we can simplify further by dividing 256 by 16:

${= \frac{16}{256}}$

${= \frac{1}{16}}$

Therefore, ${\left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4 = \frac{1}{16}}$. This example illustrates the importance of simplifying expressions before performing multiplication to avoid dealing with large numbers. By applying the rules of exponents and fraction simplification, we can efficiently arrive at the final answer. Recognizing opportunities for cancellation, such as the common factor of 81 in this case, is a key skill in simplifying mathematical expressions. Furthermore, breaking down the problem into smaller steps—calculating each exponential term separately before multiplying—helps to manage complexity and reduce errors.

m. Evaluating ${(5^{-1} - 3^{-1})^{-1}}$

This expression involves negative exponents and operations within parentheses. First, we evaluate the terms inside the parentheses. Recall that ${a^{-1} = \frac{1}{a}}$, so:

${5^{-1} = \frac{1}{5}}$

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

Substituting these values into the expression, we get:

${\left(\frac{1}{5} - \frac{1}{3}\right)^{-1}}$

To subtract the fractions, we need a common denominator, which is 15:

${= \left(\frac{3}{15} - \frac{5}{15}\right)^{-1}}$

${= \left(\frac{3-5}{15}\right)^{-1}}$

${= \left(\frac{-2}{15}\right)^{-1}}$

Now, we apply the negative exponent rule again, which means taking the reciprocal:

${= \frac{1}{\frac{-2}{15}}}$

${= \frac{15}{-2}}$

${= -\frac{15}{2}}$

Thus, ${(5^{-1} - 3^{-1})^{-1} = -\frac{15}{2}}$. This problem demonstrates the order of operations (PEMDAS/BODMAS) and the importance of handling negative exponents correctly. First, we addressed the expressions inside the parentheses, which involved finding a common denominator to subtract fractions. Then, we applied the rule for negative exponents, which states that ${a^{-1}}$ is the reciprocal of ${a}$. This step is crucial for simplifying the expression. Finally, we simplified the resulting fraction to arrive at the answer. This type of problem reinforces the interconnectedness of various mathematical concepts and the need for a systematic approach to problem-solving.

In this comprehensive guide, we have explored the evaluation of various mathematical expressions involving exponents, including negative and zero exponents. We systematically broke down each problem, providing detailed explanations and step-by-step solutions. From understanding the basic definitions of exponents to applying rules for negative exponents and zero exponents, this exploration has equipped you with the essential tools to confidently tackle a wide range of mathematical challenges. Mastering these concepts is crucial for building a solid foundation in mathematics and for success in more advanced topics. By practicing these types of problems and understanding the underlying principles, you can develop a strong command of exponential operations.