Evaluating Expressions With Exponents Zero Exponents And Negative Exponents

This article delves into the evaluation of various mathematical expressions involving exponents, negative exponents, and zero exponents. We will explore the fundamental rules governing these operations and apply them to solve a series of problems. Understanding these concepts is crucial for building a strong foundation in algebra and higher-level mathematics. Let's embark on this journey of mathematical exploration.

Evaluating Expressions with Exponents

Exponents are a concise way to represent repeated multiplication. For example, ${x^n}$ means multiplying ${x}$ by itself ${n}$ times. The number ${x}$ is called the base, and ${n}$ is the exponent or power. When dealing with fractions raised to a power, the exponent applies to both the numerator and the denominator. Our first expression, ${\left(\frac{-7}{10}\right)^3}$, falls into this category. To evaluate this, we need to cube both -7 and 10. This means we multiply -7 by itself three times (-7 * -7 * -7) and 10 by itself three times (10 * 10 * 10). Remembering that a negative number raised to an odd power results in a negative number, we can proceed with the calculation. The result will be a fraction with a negative numerator and a positive denominator, representing a negative rational number. This concept of distributing the exponent over the numerator and denominator is a cornerstone of exponent manipulation and is essential for simplifying more complex expressions involving rational bases.

Solving (\left(\frac{-7}{10}

ight)^3)

To solve ${\left(\frac{-7}{10}\right)^3}$, we apply the exponent to both the numerator and the denominator:

${ \left(\frac{-7}{10}\right)^3 = \frac{(-7)^3}{10^3} = \frac{-7 \times -7 \times -7}{10 \times 10 \times 10} = \frac{-343}{1000} }$

Thus, ${\left(\frac{-7}{10}\right)^3 = -\frac{343}{1000}}$. This example illustrates a fundamental rule of exponents: when a fraction is raised to a power, both the numerator and the denominator are raised to that power. This principle is applicable to any fraction and any integer exponent. Furthermore, the sign of the result is determined by the sign of the base and whether the exponent is even or odd. A negative base raised to an odd power results in a negative number, while a negative base raised to an even power yields a positive number. These rules are crucial for accurate calculations involving exponents.

Understanding Zero Exponents

Any non-zero number raised to the power of 0 equals 1. This is a fundamental rule in exponents and is often used in simplifying expressions. The expression ${65(47^0 + 33^0)}$ demonstrates this rule. Here, we have two terms, ${47^0}$ and ${33^0}$, both raised to the power of 0. According to the rule, each of these terms equals 1. Therefore, the expression simplifies to 65 multiplied by the sum of 1 and 1. This highlights the importance of understanding and applying the zero exponent rule correctly. It is a common point of confusion for students, but its consistent application is essential for accurate mathematical manipulations. The zero exponent rule is not just an isolated concept; it is intricately linked to other exponent rules, such as the quotient rule and the negative exponent rule, and plays a vital role in maintaining consistency and coherence in mathematical operations.

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

To evaluate ${65(47^0 + 33^0)}$, we use the rule that any non-zero number raised to the power of 0 is 1:

${ 65(47^0 + 33^0) = 65(1 + 1) = 65(2) = 130 }$

Therefore, ${65(47^0 + 33^0) = 130}$. This calculation exemplifies the direct application of the zero exponent rule. It's important to remember that this rule applies to any non-zero base. The logic behind this rule can be understood through the properties of exponents and division. For instance, consider ${x^n / x^n}$. According to the quotient rule of exponents, this is equal to ${x^{n-n} = x^0}$. However, any number divided by itself is 1. Therefore, ${x^0}$ must equal 1. This understanding helps solidify the concept rather than just memorizing the rule.

Working with Negative Exponents

A negative exponent indicates the reciprocal of the base raised to the positive value of the exponent. For example, ${x^{-n} = \frac{1}{x^n}}$. In the expression ( \left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2}), we have negative exponents. To simplify this, we first need to understand how to handle division with exponents. The rule states that when dividing powers with the same base, we subtract the exponents. So, the expression becomes ( \left(\frac{15}{47}\right)^{-4 - (-2)}), which simplifies to ( \left(\frac{15}{47}\right)^{-2}). Now, we apply the negative exponent rule, which means we take the reciprocal of the base and raise it to the positive exponent. This transformation allows us to work with positive exponents, making the calculation more straightforward. The concept of negative exponents is crucial for representing very small numbers and is widely used in scientific notation and various fields of science and engineering.

Simplifying (

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

To simplify ( \left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2}), we use the quotient rule of exponents, which states that when dividing powers with the same base, we subtract the exponents:

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

Now, we apply the negative exponent rule, which states that ${x^{-n} = \frac{1}{x^n}}$:

${ \left(\frac{15}{47}\right)^{-2} = \left(\frac{47}{15}\right)^{2} = \frac{47^2}{15^2} = \frac{2209}{225} }$

Thus, ( \left(\frac{15}{47}\right)^{-4} \div \left(\frac{15}{47}\right)^{-2} = \frac{2209}{225}). This example demonstrates the combined use of the quotient rule and the negative exponent rule. It is essential to understand the order of operations when simplifying such expressions. First, we apply the quotient rule to combine the exponents, and then we apply the negative exponent rule to express the result with a positive exponent. This systematic approach is crucial for accurate simplification of exponential expressions.

Combining Exponent Rules

Many expressions require the application of multiple exponent rules. Consider ( \left(\frac9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4). Here, we have two fractions raised to different powers, and we need to multiply them. First, we apply the power to both the numerator and denominator of each fraction. This gives us ( \frac{9^2}{162} \times \frac{2^4}{34}). Next, we evaluate the powers ( \frac{81{256} \times \frac{16}{81}). Now we multiply the fractions. Before directly multiplying, it's beneficial to look for opportunities to simplify. We can see that 81 appears in both the numerator and denominator, so we can cancel them out. Also, 16 is a factor of 256, so we can simplify further. This step-by-step approach of applying exponent rules and simplifying is crucial for solving more complex problems efficiently and accurately. The ability to recognize patterns and simplify before performing large calculations is a hallmark of mathematical proficiency.

Simplifying (

\left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4)

To simplify ( \left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4), we first apply the power to both the numerator and denominator of each fraction:

${ \left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4 = \frac{9^2}{16^2} \times \frac{2^4}{3^4} = \frac{81}{256} \times \frac{16}{81} }$

Now, we multiply the fractions and simplify:

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

We can cancel out the 81 in the numerator and denominator:

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

Since 16 is a factor of 256, we can simplify further:

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

Thus, ( \left(\frac{9}{16}\right)^2 \times \left(\frac{2}{3}\right)^4 = \frac{1}{16}). This example showcases the importance of simplifying expressions before performing complex calculations. By recognizing common factors and canceling them out, we can significantly reduce the complexity of the problem and arrive at the solution more efficiently. This skill is particularly valuable in algebra and calculus, where expressions can often be simplified before applying more advanced techniques.

Dealing with Parentheses and Negative Exponents

The expression ${(5^{-1} - 3^{-1})^{-1}}$ involves parentheses, negative exponents, and subtraction. To evaluate this, we first address the negative exponents inside the parentheses. Recall that ${x^{-1} = \frac{1}{x}}$. So, ${5^{-1}}$ becomes ${\frac{1}{5}}$ and ${3^{-1}}$ becomes ${\frac{1}{3}}$. Now we have ( \left(\frac{1}{5} - \frac{1}{3}\right)^{-1}). Next, we perform the subtraction inside the parentheses. To subtract fractions, we need a common denominator, which in this case is 15. So, we rewrite the fractions as ( \frac{3}{15} - \frac{5}{15}), which equals ${-\frac{2}{15}}$. Now the expression is ( \left(-\frac{2}{15}\right)^{-1}). Finally, we apply the negative exponent rule again, taking the reciprocal of the fraction. Remember that the reciprocal of a negative fraction is also negative. This comprehensive approach, addressing each component of the expression systematically, is essential for tackling more complex mathematical problems. The ability to break down a problem into smaller, manageable steps is a critical skill in mathematics.

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

To evaluate ${(5^{-1} - 3^{-1})^{-1}}$, we first address the negative exponents inside the parentheses. Recall that ${x^{-1} = \frac{1}{x}}$:

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

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

Now, we substitute these values back into the expression:

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

Next, we perform the subtraction inside the parentheses. To subtract fractions, we need a common denominator, which is 15:

${ \frac{1}{5} - \frac{1}{3} = \frac{3}{15} - \frac{5}{15} = -\frac{2}{15} }$

So the expression becomes:

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

Finally, we apply the negative exponent rule again, taking the reciprocal of the fraction:

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

Thus, ${(5^{-1} - 3^{-1})^{-1} = -\frac{15}{2}}$. This example illustrates the importance of following the order of operations and applying the exponent rules systematically. We first dealt with the inner expressions, then performed the subtraction, and finally applied the outer exponent. This methodical approach ensures accurate results, especially in complex expressions involving multiple operations.

Conclusion

In this article, we have explored various mathematical expressions involving exponents, negative exponents, and zero exponents. We have demonstrated the application of fundamental rules such as the product rule, quotient rule, power rule, zero exponent rule, and negative exponent rule. Through detailed examples, we have shown how to evaluate and simplify expressions efficiently and accurately. Mastering these concepts is crucial for success in algebra and higher-level mathematics. By understanding the underlying principles and practicing regularly, you can build a strong foundation in exponent manipulation and enhance your problem-solving skills.