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Chapter 5: Problem 32

Use properties of exponents to simplify each expression. Express answers in exponential form with positive exponents only. Assume that any variables in denominators are not equal to zero. \(\left(x^{2} \cdot x^{4}\right)^{-3}\)

Short Answer

Expert verified
The simplified form of \( \left(x^{2} \cdot x^{4}\right)^{-3} \) is \( \frac{1}{x^{18}} \).

Step by step solution

01

utilise the product of powers property

The product of powers property states that when multiplying like bases, you simply add the exponents. Therefore, \( x^{2} \cdot x^{4} \) can be rewritten as \( x^{2+4} \) or \( x^{6} \).
02

apply the power of a power rule

The power of a power rule states that to raise a power to a power, you multiply the exponents. Therefore, \( \left(x^{6}\right)^{-3} \) can be rewritten as \( x^{6 \cdot -3} \) or \( x^{-18} \).
03

use the rule of negative exponents

The negative exponent rule states that a negative exponent is equivalent to the reciprocal of that amount with a positive exponent. So, \( x^{-18} \) can be rewritten as \( \frac{1}{x^{18}} \), meaning the answer is in exponential form with positive exponents only.

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Most popular questions from this chapter

Find the indicated term for the geometric sequence with first term, \(a_{1}\), and common ratio, \(r\). Find \(a_{20}\), when \(a_{1}=2, r=2\).

Determine whether each sequence is arithmetic or geometric. Then find the next two terms. \(6,3, \frac{3}{2}, \frac{3}{4}, \ldots\)

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=-8, r=-5\)

Write the first six terms of the geometric sequence with the first term, \(a_{1}\), and common ratio, \(r\). \(a_{1}=-4, r=-2\)

Write a formula for the general term (the nth term) of each arithmetic sequence. Then use the formula for \(a_{n}\) to find \(a_{20}\), the 20 th term of the sequence. \(a_{1}=9, d=2\)
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