11. 5+ log, (n + 1) = 8

Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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11. Solve, if necessary round to the nearest hundred
The image displays a mathematical equation, which is problem number 11:

\[ 5 + \log_3(n + 1) = 8 \]

This equation involves solving for the variable \( n \) where a logarithmic expression with base 3 is used. Let's break it down step-by-step for better understanding:

1. **Understanding Logarithms**: The equation includes a logarithm base 3, written as \(\log_3(n + 1)\). This represents the power to which the base 3 must be raised to yield \( n + 1 \).

2. **Isolating the Logarithmic Term**: To solve for \( n \), first isolate the logarithmic term:
   \[ \log_3(n + 1) = 8 - 5 \]
   \[ \log_3(n + 1) = 3 \]

3. **Converting to Exponential Form**: Convert the logarithmic equation to its equivalent exponential form to solve for \( n + 1 \):
   \[ n + 1 = 3^3 \]

4. **Solving the Exponential**: Calculate \( 3^3 \), which equals 27:
   \[ n + 1 = 27 \]

5. **Finding \( n \)**: Subtract 1 from both sides to solve for \( n \):
   \[ n = 27 - 1 \]
   \[ n = 26 \]

Thus, the solution to the equation is \( n = 26 \).
Transcribed Image Text:The image displays a mathematical equation, which is problem number 11: \[ 5 + \log_3(n + 1) = 8 \] This equation involves solving for the variable \( n \) where a logarithmic expression with base 3 is used. Let's break it down step-by-step for better understanding: 1. **Understanding Logarithms**: The equation includes a logarithm base 3, written as \(\log_3(n + 1)\). This represents the power to which the base 3 must be raised to yield \( n + 1 \). 2. **Isolating the Logarithmic Term**: To solve for \( n \), first isolate the logarithmic term: \[ \log_3(n + 1) = 8 - 5 \] \[ \log_3(n + 1) = 3 \] 3. **Converting to Exponential Form**: Convert the logarithmic equation to its equivalent exponential form to solve for \( n + 1 \): \[ n + 1 = 3^3 \] 4. **Solving the Exponential**: Calculate \( 3^3 \), which equals 27: \[ n + 1 = 27 \] 5. **Finding \( n \)**: Subtract 1 from both sides to solve for \( n \): \[ n = 27 - 1 \] \[ n = 26 \] Thus, the solution to the equation is \( n = 26 \).
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