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...
Related questions
Question
![**Proof of Logarithmic Identity**
We are given the equation to prove:
\[
\log_5 100 = 2(1 + \log_5 2)
\]
### Steps to Prove:
1. **Express 100 as Powers of Smaller Numbers:**
- Noticing that \(100 = 10^2\), we can rewrite \(100\) in terms of its prime factors or smaller numbers. Here, we recognize that:
\[
100 = (10)^2 = (2 \times 5)^2 = 2^2 \times 5^2
\]
2. **Apply Logarithmic Properties:**
- Using the properties of logarithms, \( \log_b (mn) = \log_b m + \log_b n \) and \( \log_b (m^n) = n\log_b m \), we can transform the original expression:
\[
\log_5 100 = \log_5 (2^2 \times 5^2) = \log_5 (2^2) + \log_5 (5^2)
\]
\[
= 2 \log_5 2 + 2 \log_5 5
\]
3. **Simplify \(\log_5 5\):**
- We know \( \log_b b = 1 \). Therefore:
\[
\log_5 5 = 1
\]
4. **Substitute Back into the Expression:**
- Replace \(\log_5 5\) in the expanded expression:
\[
\log_5 100 = 2 \log_5 2 + 2
\]
5. **Factor Out the 2:**
- Factor the 2 from the expression:
\[
= 2(\log_5 2 + 1)
\]
6. **Prove the Identity:**
- Finally, compare the given equation with our result:
\[
\log_5 100 = 2(1 + \log_5 2)
\]
- Therefore, the identity is proven to be true.
This concludes the proof that \(\log_5 100 = 2(1 + \log_5 2)\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6ad60fb3-d7c7-4aa0-8ac8-5f8ba0f10fe4%2F0a8503cd-0df9-4022-9795-d0813c2c9018%2Fia0m21_processed.png&w=3840&q=75)
Transcribed Image Text:**Proof of Logarithmic Identity**
We are given the equation to prove:
\[
\log_5 100 = 2(1 + \log_5 2)
\]
### Steps to Prove:
1. **Express 100 as Powers of Smaller Numbers:**
- Noticing that \(100 = 10^2\), we can rewrite \(100\) in terms of its prime factors or smaller numbers. Here, we recognize that:
\[
100 = (10)^2 = (2 \times 5)^2 = 2^2 \times 5^2
\]
2. **Apply Logarithmic Properties:**
- Using the properties of logarithms, \( \log_b (mn) = \log_b m + \log_b n \) and \( \log_b (m^n) = n\log_b m \), we can transform the original expression:
\[
\log_5 100 = \log_5 (2^2 \times 5^2) = \log_5 (2^2) + \log_5 (5^2)
\]
\[
= 2 \log_5 2 + 2 \log_5 5
\]
3. **Simplify \(\log_5 5\):**
- We know \( \log_b b = 1 \). Therefore:
\[
\log_5 5 = 1
\]
4. **Substitute Back into the Expression:**
- Replace \(\log_5 5\) in the expanded expression:
\[
\log_5 100 = 2 \log_5 2 + 2
\]
5. **Factor Out the 2:**
- Factor the 2 from the expression:
\[
= 2(\log_5 2 + 1)
\]
6. **Prove the Identity:**
- Finally, compare the given equation with our result:
\[
\log_5 100 = 2(1 + \log_5 2)
\]
- Therefore, the identity is proven to be true.
This concludes the proof that \(\log_5 100 = 2(1 + \log_5 2)\).
Expert Solution
Step 1: To Prove
We have to prove that
Step by step
Solved in 2 steps
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning