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Chapter 3: Problem 78

Suppose \(a\) and \(b\) are positive numbers, with \(a \neq 1\) and \(b \neq 1 .\) Show that $$ \log _{a} b=\frac{1}{\log _{b} a} $$

Short Answer

Expert verified
By introducing a variable c such that \(b = a^{c}\), we can rewrite and simplify the equation using logarithmic properties to prove the logarithmic identity \(\log _{a} b=\frac{1}{\log _{b} a}\) for the given positive numbers a and b, where \(a \neq 1\) and \(b \neq 1\).

Step by step solution

01

Introduce variable c

Introduce a variable c such that \(b = a^{c}\). This will allow us to use the properties of logarithms and manipulate the equation more effectively.
02

Write the equation using logarithmic properties

Using the definition of logarithm with the introduced variable c, we can rewrite the given equation as \[\log_a (a^c) = \frac{1}{\log_{b} a}\]
03

Simplify the left side of the equation

Using the property of logarithms, \(\log_{a}(a^c) = c\), we can simplify the left side of the equation as \[c = \frac{1}{\log_{b} a}\]
04

Write back the equation with b

In Step 1, we introduced the variable c such that \(b = a^{c}\). Now, let's rewrite the equation in terms of b by replacing c: \[\log_a b = \frac{1}{\log_{b} a}\]
05

Conclude

We have successfully proven the logarithmic identity for the given positive numbers a and b such that \(a \neq 1\) and \(b \neq 1\): \[\log _{a} b=\frac{1}{\log _{b} a}\]

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