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Understanding Logarithm Change of Base
Sep 21, 2024
Logarithm Change of Base Formula and Solving Exponential Equations
Introduction
Instructor: Jason (Math and Science Comm)
Lesson Title: Logarithm Change of Base Formula and Solving Exponential and Logarithmic Equations
Importance of Logarithms
Primarily used in solving equations, especially exponential equations.
Logarithms can be complex with different bases (e.g., base 2, 3, 5, etc.).
Difficulty arises when needing numerical answers and calculators lack specific base buttons.
Change of Base Formula
A tool that allows computation of logarithms of any base using more common logarithms (base 10 or base e).
Formula:
[ \log_a(x) = \frac{\log_b(x)}{\log_b(a)} ]
Where:
( a ) is the original base,
( b ) is the new base,
( x ) is the number being logged.
Understanding the Formula
Base on the left can be changed to any base on the right.
Example:
Change ( \log_3(9) ) to ( \frac{\log_2(9)}{\log_2(3)} ).
Can also change to base 10 or any other base.
Practical Examples
Example 1: Logarithm Calculation
To calculate ( \log_4(7) ):
Use the change of base:
[ \log_4(7) = \frac{\log_{10}(7)}{\log_{10}(4)} ]
Result: ( \approx 1.404 ) (rounded)
Example 2: Exponential Equation
For ( 3^{2x} = 5 ):
Take logarithm of both sides:
[ 2x = \log_3(5) ]
Use change of base:
[ x = \frac{\log_{10}(5)}{2 \cdot \log_{10}(3)} ]
Calculate: ( x \approx 0.7326 ) (rounded)
Additional Examples
Example:
( \log_{10}(x) = 0.8531853 )
Solve: ( x = 10^{0.8531853} \approx 7.13 ) (rounded)
Example:
( 3^x = 30 )
Solve:
[ x = \log_3(30) \approx \frac{\log_{10}(30)}{\log_{10}(3)} \approx 3.096 ]
Derivation of Change of Base Formula
Start with: ( \log_a(X) = Y ) implies ( a^Y = X ).
Take logarithm of both sides with base ( B ):
[ \log_B(a^Y) = \log_B(X) ]
Rearranging gives:
[ Y \cdot \log_B(a) = \log_B(X) ]
Solve for ( \log_a(X) ):
[ \log_a(X) = \frac{\log_B(X)}{\log_B(a)} ]
Conclusion
The change of base formula is crucial for solving logarithmic equations, especially with given bases unavailable on calculators.
Understanding how to manipulate and derive the formula enhances problem-solving skills in algebra and calculus.
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