Solve: log2 (x+9) — log2 x = 1.

Ch. 8 - Sequences, Induction, and Probability

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Blitzer 8th Edition

Ch. 8 - Sequences, Induction, and Probability

문제 87

9장, 문제 87

Solve: log₂ (x+9) — log₂ x = 1.

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Recall the logarithmic property that states: \(\log_a b - \log_a c = \log_a \left( \frac{b}{c} \right)\), which allows us to combine the difference of two logs with the same base into a single log of a quotient.

Apply this property to the equation \(\log_2 (x+9) - \log_2 x = 1\) to rewrite it as \(\log_2 \left( \frac{x+9}{x} \right) = 1\).

Rewrite the logarithmic equation in its equivalent exponential form: \(\frac{x+9}{x} = 2^1\).

Simplify the right side to get \(\frac{x+9}{x} = 2\), then multiply both sides by \(x\) to clear the denominator, resulting in \(x + 9 = 2x\).

Solve the linear equation \(x + 9 = 2x\) by isolating \(x\), and then check the solution to ensure it does not make any logarithm argument negative or zero.

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Properties of Logarithms

Logarithmic properties, such as the subtraction rule log_b(A) - log_b(B) = log_b(A/B), allow simplification of expressions involving logs. This property is essential for combining or breaking down logarithmic terms to solve equations.

Definition of Logarithms

A logarithm log_b(A) answers the question: to what power must the base b be raised to get A? Understanding this helps convert logarithmic equations into exponential form, which is often easier to solve.

Domain Restrictions in Logarithmic Functions

The argument of a logarithm must be positive, so expressions inside logs (like x and x+9) must be greater than zero. Recognizing domain restrictions ensures solutions are valid within the function's domain.