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Ch. 4 - Exponential and Logarithmic Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 5, Problem 87

Solve each logarithmic equation in Exercises 49–92. Be sure to reject any value of x that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. log x+log(x+3)=log 10

Verified step by step guidance
1
Recall the logarithmic property that states \( \log a + \log b = \log (a \times b) \). Use this to combine the left side of the equation: \( \log x + \log (x+3) = \log [x(x+3)] \).
Rewrite the equation using the combined logarithm: \( \log [x(x+3)] = \log 10 \).
Since the logarithms are equal and have the same base, set their arguments equal: \( x(x+3) = 10 \).
Expand and rewrite the equation as a quadratic: \( x^2 + 3x = 10 \), then bring all terms to one side to get \( x^2 + 3x - 10 = 0 \).
Solve the quadratic equation using the quadratic formula or factoring, then check each solution to ensure it makes the arguments of the original logarithms positive (i.e., \( x > 0 \) and \( x+3 > 0 \)) to determine valid solutions.

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Key Concepts

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

Properties of Logarithms

Understanding the properties of logarithms, especially the product rule, is essential. The product rule states that log(a) + log(b) = log(ab), which allows combining multiple logarithmic terms into a single logarithm, simplifying the equation for easier solving.
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Domain of Logarithmic Functions

The domain of a logarithmic function includes only positive real numbers. When solving logarithmic equations, it is crucial to check that the solutions do not make any argument of the logarithm zero or negative, as these values are not valid in the original equation.
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Solving Logarithmic Equations

After applying logarithmic properties, the equation often reduces to an algebraic form. Solving this algebraic equation involves isolating the variable and finding exact solutions, which can then be approximated using a calculator if needed.
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