IB Mathematics AA SL Exponents and logarithms Study Notes
IB Mathematics AA SL Exponents and logarithms Study Notes Offer a clear explanation of Exponents and logarithms , including various formula, rules, exam style questions as example to explain the topics. Worked Out examples and common problem types provided here will be sufficient to cover for topic Exponents and logarithms
Laws of Logarithms
Laws of Exponents with Rational Exponents
Rational exponents represent both roots and powers. A rational exponent like \( a^{\frac{m}{n}} \) means the nth root of \( a^m \), or equivalently,
\( a^{\frac{m}{n}} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m \)
| Exponent Rules | |
|---|---|
| For \( a \ne 0, b \ne 0 \) | |
| Product Rule | \( a^r \cdot a^s = a^{r+s} \) |
| Quotient Rule | \( \frac{a^r}{a^s} = a^{r – s} \), \( a \ne 0 \) |
| Power of a Power | \( (a^r)^s = a^{r \cdot s} \) |
| Power of a Product | \( (ab)^r = a^r \cdot b^r \) |
| Power of a Quotient | \( \left(\frac{a}{b}\right)^r = \frac{a^r}{b^r} \), \( b \ne 0 \) |
| Zero Exponent | \( a^0 = 1 \), \( a \ne 0 \) |
| Negative Exponent | \( a^{-r} = \frac{1}{a^r} \), \( a \ne 0 \) |
Examples
- Evaluate \( 8^{\frac{2}{3}} \)
- Simplify \( (27x^3)^{\frac{2}{3}} \)
- Simplify \( \frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}}} \)
- Rewrite \( \frac{1}{x^{\frac{5}{2}}} \) using a negative exponent
▶️ Answer/Explanation
- \( 8^{\frac{2}{3}} = (\sqrt[3]{8})^2 = 2^2 = 4 \)
- \( (27x^3)^{\frac{2}{3}} = 27^{\frac{2}{3}} \cdot x^{2} = (\sqrt[3]{27})^2 \cdot x^2 = 3^2 \cdot x^2 = 9x^2 \)
- \( \frac{16^{\frac{3}{4}}}{4^{\frac{1}{2}}} = \frac{(\sqrt[4]{16})^3}{\sqrt{4}} = \frac{2^3}{2} = \frac{8}{2} = 4 \)
- \( \frac{1}{x^{\frac{5}{2}}} = x^{-\frac{5}{2}} \)
Laws of Logarithms
Laws of Logarithms
Logarithms are the inverses of exponents. If \( a^x = b \), then \( \log_a b = x \). The following rules apply for all logarithms, where \( a > 0 \), \( a \ne 1 \), and \( x, y > 0 \):
| Key Laws of Logarithms | |
|---|---|
| Product Rule | \( \log_a(xy) = \log_a x + \log_a y \) |
| Quotient Rule | \( \log_a\left(\frac{x}{y}\right) = \log_a x – \log_a y \) |
| Power Rule | \( \log_a(x^r) = r \log_a x \) |
| Log of 1 | \( \log_a 1 = 0 \) |
| Log of the base | \( \log_a a = 1 \) |
Examples
- Simplify \( \log_{10}(1000) \)
- Simplify \( \log_2(8 \times 4) \)
- Evaluate \( \log_3\left(\frac{27}{3}\right) \)
- Simplify \( \log_5(25^3) \)
- Convert \( \log_2 7 \) into base 10
▶️ Answer/Explanations
- \( \log_{10}(1000) = \log_{10}(10^3) = 3 \)
- \( \log_2(8 \times 4) = \log_2 8 + \log_2 4 = 3 + 2 = 5 \)
- \( \log_3\left(\frac{27}{3}\right) = \log_3 27 – \log_3 3 = 3 – 1 = 2 \)
- \( \log_5(25^3) = \log_5((5^2)^3) = \log_5(5^6) = 6 \)
- \( \log_2 7 = \frac{\log_{10} 7}{\log_{10} 2} \approx \frac{0.8451}{0.3010} \approx 2.807 \)
Change of Base of a Logarithm
Change of Base of a Logarithm
Most calculators only support logarithms with base 10 (common log) or base \( e \) (natural log). To evaluate logs with other bases, we use the Change of Base Formula:
\( \log_b x = \frac{\log_k x}{\log_k b} \)
Where:
- \( b \): the original base
- \( x \): the argument of the logarithm
- \( k \): any other base (commonly 10 or \( e \))
This allows you to compute any logarithm using a calculator that has only log (base 10) or ln (base \( e \)) functions.
Examples
- Evaluate \( \log_2 50 \) using the change of base formula and a calculator.
- Evaluate \( \log_7 100 \) using base 10 logarithms.
- Evaluate \( \log_3 81 \) using base \( e \) (ln).
▶️ Answer/Explanation
- \( \log_2 50 = \frac{\log_{10} 50}{\log_{10} 2} \approx \frac{1.69897}{0.30103} \approx 5.644 \)
- \( \log_7 100 = \frac{\log_{10} 100}{\log_{10} 7} = \frac{2}{0.8451} \approx 2.366 \)
- \( \log_3 81 = \frac{\ln 81}{\ln 3} = \frac{4.3944}{1.0986} \approx 4 \)
Solving Exponential Equations Using Logarithms
Solving Exponential Equations Using Logarithms
Exponential equations involve unknowns in the exponent, such as \( a^x = b \). When you can’t rewrite both sides with the same base, you apply logarithms to both sides to bring the exponent down and solve.
Steps:
- Isolate the exponential expression (if necessary)
- Take logs of both sides (common log or natural log)
- Use the power rule: \( \log(a^x) = x \log(a) \)
- Solve the resulting equation algebraically
Graphical Link: Exponential and logarithmic graphs are reflections of each other across the line \( y = x \). Solving \( a^x = b \) is equivalent to finding where the graph of \( y = a^x \) intersects \( y = b \).
Examples
- Solve \( 2^x = 10 \)
- Solve \( 5^{2x – 1} = 100 \)
- Solve \( e^{x + 2} = 7 \)
▶️ Answer/Explanation
- Take log of both sides: \( \log(2^x) = \log(10) \)
Use power rule: \( x \log(2) = \log(10) \)
\( x = \frac{\log(10)}{\log(2)} = \frac{1}{0.3010} \approx 3.322 \) - Take log of both sides: \( \log(5^{2x – 1}) = \log(100) \)
\( (2x – 1) \log(5) = \log(100) \)
\( (2x – 1) = \frac{2}{\log(5)} = \frac{2}{0.6990} \approx 2.861 \)
Solve: \( 2x = 3.861 \Rightarrow x \approx 1.931 \) - Take natural log (ln): \( \ln(e^{x+2}) = \ln(7) \)
\( x + 2 = \ln(7) \approx 1.9459 \)
\( x = 1.9459 – 2 = -0.0541 \)
Solving Exponential Equations Using a GDC
Graphical Display Calculators (GDCs) can be used to solve exponential equations numerically or visually by finding points of intersection between two graphs.
Method:
- Enter both sides of the equation as separate functions in the GDC
- Use the graphing mode
- Find the point of intersection (where the graphs are equal)
- The x-coordinate of the intersection point is the solution
Examples(GDC)
- Solve \( 3^x = 20 \) using a GDC
- Solve \( 2^{2x+1} = 50 \) using graph intersection
- Solve \( e^x = 5 \) using a calculator
▶️ Answer/Explanation
- On the GDC:
- Graph \( y_1 = 3^x \)
- Graph \( y_2 = 20 \)
- Use “Intersect” function
- On the GDC:
- Graph \( y_1 = 2^{2x+1} \)
- Graph \( y_2 = 50 \)
- Use “Intersect” to find the solution
- On the GDC:
- Use the natural log function: \( x = \ln(5) \)