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Ch. 4 - Inverse, Exponential, and Logarithmic Functions
Lial - College Algebra 13th Edition
Lial13th EditionCollege AlgebraISBN: 9780136881063Not the one you use?Change textbook
All textbooksLial 13th EditionCh. 4 - Inverse, Exponential, and Logarithmic FunctionsProblem 95
Chapter 5, Problem 95

Solve each equation. See Examples 4–6. 1/27 = x-3

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Start with the given equation: \(\frac{1}{27} = x^{-3}\).
Recall that a negative exponent means the reciprocal, so rewrite \(x^{-3}\) as \(\frac{1}{x^3}\), giving \(\frac{1}{27} = \frac{1}{x^3}\).
Since the fractions are equal and both have numerator 1, set the denominators equal: \(27 = x^3\).
To solve for \(x\), take the cube root of both sides: \(x = \sqrt[3]{27}\).
Simplify the cube root to find the value of \(x\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Negative Exponents

A negative exponent indicates the reciprocal of the base raised to the corresponding positive exponent. For example, x^(-n) equals 1 divided by x^n. Understanding this allows rewriting expressions like x^-3 as 1/x^3.
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Properties of Exponents

Exponent rules, such as a^(m) * a^(n) = a^(m+n) and (a^m)^n = a^(m*n), help simplify and solve equations involving powers. Applying these properties enables manipulation of expressions to isolate variables.
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Solving Exponential Equations

Solving equations with variables in exponents often involves rewriting both sides with the same base or using logarithms. In this problem, expressing both sides as powers of 3 allows equating exponents to find x.
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