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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 5: Q 109. (page 309)

Show that $\log_{a} (\frac{M}{N}) = \log_{a} M - \log_{a} N,$ where a, M, and N are positive real numbers and $a 鈮� 1$ .

Short Answer

Expert verified

It is proved that $\log_{a} (\frac{M}{N}) = \log_{a} M - \log_{b} N$ where a, M, and N are positive real numbers and $a 鈮� 1$ .

Step by step solution

01

Step 1. Applying law of Exponents.

Let $A = \log_{a} M & B = \log_{a} N$ . These expressions are equivalent to exponential expressions $a^{M} & a^{B} = N$ .

To prove $\log_{a} (\frac{M}{N}) = \log_{a} M - \log_{a} N$ .

Consider the Left hand side and substitute $a^{A}$ for M and $a^{B}$ for N.

$\log_{a} (\frac{M}{N}) = \log_{a} (\frac{a^{A}}{a^{B}}) = \log_{a} (a^{A} 脳 a^{- B}) = \log_{a} (a^{A - B}) (Using law of exponents)$

02

Step 2. Applying the property of logarithms : logaar=r.

Applying the property of logarithms in the previous step.

\log_{a} (\frac{M}{N}) = \log_{a} (a^{A - B}) = A - B = \log_{a} M - \log_{b} N (S u b s t i t u t i n g A = \log_{a} M & B = \log_{a} N)

Hence it is proved that $\log_{a} (\frac{M}{N}) = \log_{a} M - \log_{a} N$ .

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Most popular questions from this chapter

True or False. The domain of the composite function $(f 鈭� g) (x)$ is the same as the domain of $g (x)$ .

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