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. 6(x+1) = 4(2x-1)
Table of contents
- 0. Review of Algebra4h 18m
- 1. Equations & Inequalities3h 18m
- 2. Graphs of Equations1h 43m
- 3. Functions2h 17m
- 4. Polynomial Functions1h 44m
- 5. Rational Functions1h 23m
- 6. Exponential & Logarithmic Functions2h 28m
- 7. Systems of Equations & Matrices4h 5m
- 8. Conic Sections2h 23m
- 9. Sequences, Series, & Induction1h 22m
- 10. Combinatorics & Probability1h 45m
6. Exponential & Logarithmic Functions
Solving Exponential and Logarithmic Equations
Problem 23
Textbook Question
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
Verified step by step guidance1
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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Rational Exponents
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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