Rewrite logg (13) as an expression that only has terms involving log10-

Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Transcription of the Educational Content:**

Title: Change of Base Formula in Logarithms

---

**Problem Statement:**

*Rewrite \(\log_8(13)\) as an expression that only has terms involving \(\log_{10}\).*

**Explanation:**

To express \(\log_8(13)\) using \(\log_{10}\), we can use the change of base formula for logarithms. The change of base formula states:

\[
\log_b(a) = \frac{\log_c(a)}{\log_c(b)}
\]

Where \(b\) is the original base, \(a\) is the argument, and \(c\) is the new base you want to use. For this problem, we want to use base \(10\). So, applying the formula:

\[
\log_8(13) = \frac{\log_{10}(13)}{\log_{10}(8)}
\]

This allows us to express \(\log_8(13)\) entirely in terms of base \(10\) logarithms.
Transcribed Image Text:**Transcription of the Educational Content:** Title: Change of Base Formula in Logarithms --- **Problem Statement:** *Rewrite \(\log_8(13)\) as an expression that only has terms involving \(\log_{10}\).* **Explanation:** To express \(\log_8(13)\) using \(\log_{10}\), we can use the change of base formula for logarithms. The change of base formula states: \[ \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \] Where \(b\) is the original base, \(a\) is the argument, and \(c\) is the new base you want to use. For this problem, we want to use base \(10\). So, applying the formula: \[ \log_8(13) = \frac{\log_{10}(13)}{\log_{10}(8)} \] This allows us to express \(\log_8(13)\) entirely in terms of base \(10\) logarithms.
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