Chapter 0: Problem 88
Simplify each expression. $$\left(\frac{8}{27}\right)^{-2 / 3}$$
Short Answer
Expert verified
The simplified expression is \(\frac{9}{4}\).
Step by step solution
01
Understanding Negative Exponents
When dealing with negative exponents, we can use the rule \(\frac{1}{a^n} = a^{-n}\). This allows us to rewrite the expression with positive exponents. So, \(\big(\frac{8}{27}\big)^{-2/3}\) can be rewritten as \(\big(\frac{27}{8}\big)^{2/3}\).
02
Applying the Rational Exponent
The exponent \(\frac{2}{3}\) can be separated into two parts: the denominator (3), which indicates a root, and the numerator (2), which indicates a power. First, apply the cube root: \(\big(\frac{27}{8}\big)^{1/3}\).
03
Calculating the Cube Root
Find the cube root of both the numerator and the denominator separately. \(\big(\frac{27}{8}\big)^{1/3} = \frac{27^{1/3}}{8^{1/3}} = \frac{3}{2}\).
04
Applying the Square
Now, raise the result \(\frac{3}{2}\) to the power of 2. \(\big(\frac{3}{2}\big)^{2} = \frac{3^2}{2^2} = \frac{9}{4}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Negative Exponents
Negative exponents can seem a bit tricky at first, but they are simply another way to express division. When we see a negative exponent, for instance, \(\frac{1}{a^n}\), it means the same as \(a^{-n}\). For example, \(5^{-3}\) is equivalent to \(\frac{1}{5^3}\). This rule helps us rewrite expressions with negative exponents into a more manageable form with positive exponents. In our exercise, \(\bigg(\frac{8}{27}\bigg)^{-2/3}\) was transformed into \(\bigg(\frac{27}{8}\bigg)^{2/3}\) by using this rule.
Breaking down expressions with negative exponents helps simplify them step by step and avoid confusion.
Breaking down expressions with negative exponents helps simplify them step by step and avoid confusion.
Rational Exponents
Rational exponents can be split into two components: the numerator and the denominator. The denominator of the rational exponent indicates the root, and the numerator indicates the power.
For instance, in \(\bigg(\frac{27}{8}\bigg)^{2/3}\), the \(2/3\) tells us to first take the cube root (because of the 3 in the denominator) and then raise the result to the power of 2 (because of the 2 in the numerator). This two-step approach makes it easier to handle complex expressions.
For instance, in \(\bigg(\frac{27}{8}\bigg)^{2/3}\), the \(2/3\) tells us to first take the cube root (because of the 3 in the denominator) and then raise the result to the power of 2 (because of the 2 in the numerator). This two-step approach makes it easier to handle complex expressions.
Cube Root
The cube root is the opposite of cubing a number; it is finding which number, when cubed, equals the initial number. For example, the cube root of 27 is 3, as \(3^3 = 27\).
In our problem, we had to find the cube root of both the numerator and the denominator. The cube root of 27 is 3, and the cube root of 8 is 2. By simplifying \(\bigg(\frac{27}{8}\bigg)^{1/3}\) to \(\frac{3}{2}\), we make the next steps much simpler to follow.
In our problem, we had to find the cube root of both the numerator and the denominator. The cube root of 27 is 3, and the cube root of 8 is 2. By simplifying \(\bigg(\frac{27}{8}\bigg)^{1/3}\) to \(\frac{3}{2}\), we make the next steps much simpler to follow.
Fractional Exponents
Fractional exponents are a way to express both roots and powers in one compact form. For example, \(a^{m/n}\) means we take the n-th root of a and then raise it to the m-th power.
In our exercise, we dealt with \(\bigg(\frac{27}{8}\bigg)^{2/3}\) by first applying the cube root to get \(\frac{3}{2}\).
Then, we raised \(\frac{3}{2}\) to the power of 2 to get the final simplified form: \(\frac{9}{4}\). This demonstrates that separating the fractional exponent into root and power makes complex expressions much more manageable.
In our exercise, we dealt with \(\bigg(\frac{27}{8}\bigg)^{2/3}\) by first applying the cube root to get \(\frac{3}{2}\).
Then, we raised \(\frac{3}{2}\) to the power of 2 to get the final simplified form: \(\frac{9}{4}\). This demonstrates that separating the fractional exponent into root and power makes complex expressions much more manageable.