IB MYP 4-5 Maths- Logarithms, including laws of logarithms - Study Notes - New Syllabus
IB MYP 4-5 Maths- Logarithms, including laws of logarithms – Study Notes
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- Logarithms, including laws of logarithms
IB MYP 4-5 Maths- Logarithms, including laws of logarithms – Study Notes – All topics
Logarithms
Logarithms
A logarithm is the inverse of an exponent. It tells us what power we must raise a base to, in order to get a given number.
Definition: If \( b^x = a \), then \( \log_b a = x \)
This reads as: “log base \( b \) of \( a \) is \( x \)”, meaning \( b \) raised to the power \( x \) gives \( a \).
Vocabulary:
- Base: The number we raise to a power.
- Argument: The number we’re taking the log of.
- Logarithmic Form: \( \log_b a = x \)
- Exponential Form: \( b^x = a \)
Laws of Logarithms
Logarithmic expressions follow certain rules, much like exponent rules:
Law | Formula | Example |
---|---|---|
Product Law | \( \log_b(xy) = \log_b x + \log_b y \) | \( \log_2(8 \cdot 4) = \log_2 8 + \log_2 4 \) |
Quotient Law | \( \log_b\left(\frac{x}{y}\right) = \log_b x – \log_b y \) | \( \log_2\left(\frac{16}{4}\right) = \log_2 16 – \log_2 4 \) |
Power Law | \( \log_b(x^n) = n \log_b x \) | \( \log_3(9^2) = 2 \log_3 9 \) |
Change of Base | \( \log_b x = \frac{\log_k x}{\log_k b} \) | \( \log_2 10 = \frac{\log_{10} 10}{\log_{10} 2} \) |
Important Properties
- \( \log_b 1 = 0 \) (because \( b^0 = 1 \))
- \( \log_b b = 1 \) (because \( b^1 = b \))
- \( \log_b b^x = x \)
- \( b^{\log_b x} = x \)
Graph of a Logarithmic Function
The function \( y = \log_b x \) is the inverse of \( y = b^x \)
- The graph passes through (1, 0)
- It has a vertical asymptote at \( x = 0 \)
- It increases slowly for larger \( x \)
- Domain: \( x > 0 \)
- Range: All real numbers (\( \mathbb{R} \))
Example base 10 graph: \( y = \log_{10} x \)
Example base e (natural log): \( y = \ln x \) – same rules but curved differently
Using Technology (Calculator)
Most calculators can compute:
- \( \log_{10} x \) (common log)
- \( \ln x = \log_e x \) (natural log)
To find \( \log_2 5 \), use:
\( \log_2 5 = \frac{\log 5}{\log 2} \approx \boxed{2.32} \)
Applications of Logarithms
- Used to solve exponential equations: \( 2^x = 17 \Rightarrow x = \log_2 17 \)
- Found in formulas involving growth/decay, earthquakes (Richter scale), sound (decibels), etc.
Example:
Solve: \( 3^x = 81 \)
▶️ Answer/Explanation
Take log on both sides:
\( \log(3^x) = \log(81) \Rightarrow x \log 3 = \log 81 \)
\( x = \frac{\log 81}{\log 3} = \boxed{4} \)
Example:
Evaluate \( \log_5 125 \) using laws of logarithms.
▶️ Answer/Explanation
Write 125 as a power of 5: \( 125 = 5^3 \)
\( \log_5 125 = \log_5(5^3) = \boxed{3} \)
Example :
Evaluate: \( \log_3 81 \)
▶️ Answer/Explanation
\( 81 = 3^4 \), so \( \log_3 81 = 4 \)
Example:
Expand: \( \log_5 (25x^2) \)
▶️ Answer/Explanation
\( \log_5 (25x^2) = \log_5 25 + \log_5 x^2 = 2 + 2\log_5 x \)
Example :
The sound level \( L \) in decibels is given by: \( L = 10 \log\left(\frac{I}{I_0}\right) \), where \( I \) is the sound intensity and \( I_0 \) is the reference level.
If \( I = 1000I_0 \), find \( L \).
▶️ Answer/Explanation
\( L = 10 \log(1000) = 10 \cdot \log(10^3) = 10 \cdot 3 = \boxed{30\ \text{dB}} \)
Example :
Solve for \( x \): \( 3^x = 50 \)
▶️ Answer/Explanation
Take log on both sides:
\( \log 3^x = \log 50 \Rightarrow x \log 3 = \log 50 \)
\( x = \frac{\log 50}{\log 3} \approx \frac{1.699}{0.477} \approx \boxed{3.56} \)