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Chapter 11: Problem 25

Evaluate the given expressions. $$125^{-2 / 3}-100^{-3 / 2}$$

Short Answer

Expert verified
The expression evaluates to \(\frac{39}{1000}\).

Step by step solution

01

Understand The Expression

The given expression is \(125^{-2/3} - 100^{-3/2}\). This expression involves negative fractional exponents.
02

Simplify with Fractional Exponents

Fractional exponents can be rewritten as roots. For example, \(a^{b/c} = \sqrt[c]{a^b}\). Therefore, \(125^{-2/3}\) becomes \(\left(125^{2/3}\right)^{-1}\) or \(\frac{1}{125^{2/3}}\). Similarly, \(100^{-3/2}\) can be rewritten as \(\frac{1}{100^{3/2}}\).
03

Calculate the Cube and Square Roots

Start by calculating \(125^{2/3}\):- First, \(125^{1/3}\) is the cube root of 125, which is 5 because \(5^3 = 125\). - Thus, \(125^{2/3} = 5^2 = 25\).Next, calculate \(100^{3/2}\):- First, \(100^{1/2}\) is the square root of 100, which is 10.- Thus, \(100^{3/2} = 10^3 = 1000\).
04

Evaluate the Expression

Using the simplified roots, the expression becomes:- \(125^{-2/3} = \frac{1}{25}\)- \(100^{-3/2} = \frac{1}{1000}\)Now subtract the two fractions:\[\frac{1}{25} - \frac{1}{1000}\]
05

Find a Common Denominator

The least common denominator of 25 and 1000 is 1000.- Convert \(\frac{1}{25}\) to \(\frac{40}{1000}\).- Subtract: \(\frac{40}{1000} - \frac{1}{1000} = \frac{39}{1000}\).
06

Conclusion

The evaluated expression \(125^{-2/3} - 100^{-3/2}\) simplifies to \(\frac{39}{1000}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fractional Exponents
Fractional exponents are a compact way to express roots and powers in a single expression. For instance, when you encounter an exponent like \(a^{b/c}\), it can be interpreted as a two-step process involving both an exponent and a root.
  • The numerator \(b\) is the power to which you raise the base \(a\).
  • The denominator \(c\) represents the root you need to take.
For example, with \(125^{-2/3}\):
  • The negative sign indicates it's a reciprocal or one over the positive power's value.
  • \(125^{2/3}\) equates to \(\sqrt[3]{125^2}\) or the cube root of \(125^2\).
  • This comprehensive concept underscores how roots and powers are connected.
Cube and Square Roots
Roots are essential mathematical operations that help in understanding powers in a nuanced way.
  • The square root \( \sqrt{a} \) finds the number which, when multiplied by itself, results in \(a\).
  • The cube root \(\sqrt[3]{a}\) seeks the number which, when used three times in a product, equals \(a\).
For example:
  • The square root of 100 is 10, as \(10 \times 10 = 100\).
  • The cube root of 125 is 5, due to \(5 \times 5 \times 5 = 125\).
Finding roots is crucial to simplify expressions that contain fractional exponents.
Common Denominator
To subtract fractions, a common denominator is key. It allows the fractions to share the same base.
  • If fractions have different denominators, convert them using the least common multiple (LCM) of the two denominators.
  • The LCM ensures they share a common denominator, making addition or subtraction straightforward.
When dealing with \(\frac{1}{25} - \frac{1}{1000}\), we use 1000 as the common denominator:
  • Convert \(\frac{1}{25}\) to \(\frac{40}{1000}\) because 25 fits into 1000 exactly 40 times.
  • This conversion allows for easy subtraction, simplifying the operation to \(\frac{39}{1000}\).
Having a common denominator facilitates arithmetic operations on fractions by aligning their bases.

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Most popular questions from this chapter

Express each of the given expressions in simplest form with only positive exponents. $$2 \times 3^{-1}+4 \times 3^{-2}$$

Perform the indicated operations, expressing answers in simplest form with rationalized denominators. Then verify the result with a calculator. $$(2 \sqrt{5}-\sqrt{7})(3 \sqrt{5}+\sqrt{7})$$

Evaluate the given expressions. $$32^{0.4}+25^{-0.5}$$

Graph the given functions. $$f(x)=2 x^{2 / 3}$$

Express each of the given expressions in simplest form with only positive exponents. $$a b\left(\frac{a^{-2}}{b^{2}}\right)^{-3}\left(\frac{a^{-3}}{b^{5}}\right)^{2}$$
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