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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
Chapter 5, Problem 23

Solve each equation. In Exercises 11–34, give irrational solutions as decimals correct to the nearest thousandth. In Exercises 35-40, give solutions in exact form. e3x-7 • e-2x = 4e

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1
Start by using the property of exponents that states when multiplying expressions with the same base, you add the exponents: \(e^{3x - 7} \cdot e^{-2x} = e^{(3x - 7) + (-2x)}\).
Simplify the exponent by combining like terms: \((3x - 7) + (-2x) = x - 7\), so the equation becomes \(e^{x - 7} = 4e\).
Rewrite the right side to have the same base \(e\): \(4e = 4 \cdot e^1 = 4e^1\).
Since the bases are the same (both are \(e\)), set the exponents equal to each other: \(x - 7 = 1 + \ln(4)\), where \(\ln(4)\) comes from expressing 4 as \(e^{\ln(4)}\).
Solve for \(x\) by isolating it: \(x = 1 + \ln(4) + 7\).

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Key Concepts

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

Properties of Exponents

Understanding how to manipulate expressions with exponents is essential. When multiplying terms with the same base, add their exponents (e.g., e^a * e^b = e^(a+b)). This property simplifies the given equation by combining exponential terms.
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Solving Exponential Equations

To solve equations involving exponentials, isolate the exponential expression and apply logarithms if necessary. This allows converting the equation into a linear form in terms of the variable, making it easier to solve for x.
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Exact vs. Approximate Solutions

The problem distinguishes between giving solutions in exact form and decimal approximations. Exact solutions involve expressions with constants like e or logarithms, while approximate solutions require numerical evaluation rounded to a specified decimal place.
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