e In 11 – log 33 ™ + log 100

Trigonometry (11th Edition)
11th Edition
ISBN:9780134217437
Author:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Publisher:Margaret L. Lial, John Hornsby, David I. Schneider, Callie Daniels
Chapter1: Trigonometric Functions
Section: Chapter Questions
Problem 1RE: 1. Give the measures of the complement and the supplement of an angle measuring 35°.
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simplify completely

The expression in the image is:

\[ e^{\ln 11} - \log_3 3^m + \log 100 \]

### Explanation:

1. **\(e^{\ln 11}\)**:
   - The expression \(e^{\ln x}\) simplifies to \(x\) due to the inverse properties of the exponential function and the natural logarithm.
   - Therefore, \(e^{\ln 11}\) simplifies to 11.

2. **\(-\log_3 3^m\)**:
   - This represents a logarithm with base 3. The expression simplifies due to the property \(\log_b b^x = x\).
   - Therefore, \(-\log_3 3^m\) simplifies to \(-m\).

3. **\(+ \log 100\)**:
   - This is a common logarithm (base 10). 
   - Since \(100 = 10^2\), \(\log 100 = \log 10^2 = 2\).

Using these simplifications, the entire expression becomes:

\[ 11 - m + 2 \]

or

\[ 13 - m \]
Transcribed Image Text:The expression in the image is: \[ e^{\ln 11} - \log_3 3^m + \log 100 \] ### Explanation: 1. **\(e^{\ln 11}\)**: - The expression \(e^{\ln x}\) simplifies to \(x\) due to the inverse properties of the exponential function and the natural logarithm. - Therefore, \(e^{\ln 11}\) simplifies to 11. 2. **\(-\log_3 3^m\)**: - This represents a logarithm with base 3. The expression simplifies due to the property \(\log_b b^x = x\). - Therefore, \(-\log_3 3^m\) simplifies to \(-m\). 3. **\(+ \log 100\)**: - This is a common logarithm (base 10). - Since \(100 = 10^2\), \(\log 100 = \log 10^2 = 2\). Using these simplifications, the entire expression becomes: \[ 11 - m + 2 \] or \[ 13 - m \]
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