The Fundamental Logarithm Definition
- Equivalence Relation: The logarithm to base $b$ of a positive number $a$ is the power to which $b$ must be raised to equal $a$:
$$\log_b(a) = c \iff b^c = a \quad (b > 0, \; b \ne 1, \; a > 0)$$
• $b$ is the base.
• $a$ is the argument (must strictly satisfy $a > 0$).
• $c$ is the exponent / power. - Immediate Identity Inferences: $$\log_b(1) = 0 \quad (\text{since } b^0 = 1)$$ $$\log_b(b) = 1 \quad (\text{since } b^1 = b)$$ $$\log_b(b^k) = k \quad (\text{logarithm cancels base } b)$$ $$b^{\log_b(a)} = a \quad (\text{exponential cancels log } b)$$
- Standard Bases in IB Mathematics:
• Common Logarithm: $\log_{10}(x)$ or $\lg(x)$ (base 10).
• Natural Logarithm: $\ln(x) = \log_e(x)$ (base $e \approx 2.71828$).
Example 1 · Criterion A
Converting Between Exponential and Logarithmic Forms
(a) Write in logarithmic form: (i) $2^6 = 64$ (ii) $5^{-3} = \frac{1}{125}$.
(b) Write in exponential form: (i) $\log_3(81) = 4$ (ii) $\ln(x) = 5$.
(c) Evaluate without a calculator: $\log_4(32)$.
- (a) Exponential to log:
(i) $2^6 = 64 \iff \log_2(64) = 6$.
(ii) $5^{-3} = \frac{1}{125} \iff \log_5\left(\frac{1}{125}\right) = -3$. - (b) Log to exponential:
(i) $\log_3(81) = 4 \iff 3^4 = 81$.
(ii) $\ln(x) = 5 \iff e^5 = x$. - (c) Evaluating $\log_4(32)$: Let $y = \log_4(32) \implies 4^y = 32$. Express both sides in common base $2$: $$(2^2)^y = 2^5 \implies 2^{2y} = 2^5 \implies 2y = 5 \implies y = \frac{5}{2} = 2.5$$
Answers: (a) (i) $\log_2(64)=6$, (ii) $\log_5(1/125)=-3$; (b) (i) $3^4=81$, (ii) $x=e^5$; (c) $\frac{5}{2}$.
The Three Operational Laws of Logarithms
- Law 1 (Product Law): The logarithm of a product is the sum of the logarithms: $$\log_b(xy) = \log_b(x) + \log_b(y) \quad (x, y > 0)$$
- Law 2 (Quotient Law): The logarithm of a quotient is the difference of the logarithms: $$\log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \quad (x, y > 0)$$
- Law 3 (Power Law): An exponent inside a logarithm factors out to the front as a multiplier: $$\log_b(x^k) = k \cdot \log_b(x) \quad (x > 0)$$
- Change of Base Formula: $$\log_b(a) = \frac{\log_c(a)}{\log_c(b)} = \frac{\ln a}{\ln b} = \frac{\log_{10} a}{\log_{10} b}$$
Example 2 · Criterion A & C
Condensing and Expanding Logarithmic Expressions
(a) Write as a single logarithm in simplest form: $$2\log_a(x) + 3\log_a(y) - \frac{1}{2}\log_a(z)$$ (b) Given that $\log_{10}(2) \approx 0.301$ and $\log_{10}(3) \approx 0.477$, find $\log_{10}(15)$ without a calculator.
- (a) Condensing using power, product, and quotient laws: $$\text{Apply Power Law: } \log_a(x^2) + \log_a(y^3) - \log_a(z^{1/2})$$ $$\text{Apply Product Law: } \log_a(x^2 y^3) - \log_a(\sqrt{z})$$ $$\text{Apply Quotient Law: } \log_a\left(\frac{x^2 y^3}{\sqrt{z}}\right)$$
- (b) Decomposing $\log_{10}(15)$: Notice that $15 = \frac{30}{2} = \frac{3 \times 10}{2}$: $$\log_{10}(15) = \log_{10}(3) + \log_{10}(10) - \log_{10}(2)$$ Since $\log_{10}(10) = 1$: $$\log_{10}(15) \approx 0.477 + 1 - 0.301 = 1.176$$
Answers: (a) $\log_a\left(\frac{x^2 y^3}{\sqrt{z}}\right)$; (b) $1.176$.
Solving Exponential Equations Using Logarithms
- When Bases Cannot Be Made Equal: For equations like $3^x = 20$, take the natural or common logarithm of both sides: $$\ln(3^x) = \ln(20) \implies x \ln(3) = \ln(20) \implies x = \frac{\ln(20)}{\ln(3)} \approx 2.73$$
- Equations with Different Bases on Both Sides: $$a^{px + q} = b^{rx + s} \implies (px + q)\ln a = (rx + s)\ln b$$ Expand, group all terms containing $x$ on one side, and factor out $x$.
Example 3 · Criterion A
Solving Different-Base Exponential Equations
Find the exact value of $x$ for the equation $2^{x+3} = 5^{2x - 1}$. Give your answer in terms of natural logarithms.
- Step 1: Take natural logarithms of both sides: $$\ln(2^{x+3}) = \ln(5^{2x-1})$$ $$(x + 3)\ln 2 = (2x - 1)\ln 5$$
- Step 2: Expand brackets: $$x\ln 2 + 3\ln 2 = 2x\ln 5 - \ln 5$$
- Step 3: Collect terms with $x$: $$3\ln 2 + \ln 5 = 2x\ln 5 - x\ln 2$$ $$\ln(2^3) + \ln 5 = x(2\ln 5 - \ln 2)$$ $$\ln(8 \times 5) = x(\ln 25 - \ln 2)$$ $$\ln(40) = x \ln\left(\frac{25}{2}\right)$$
- Step 4: Solve for $x$: $$x = \frac{\ln(40)}{\ln(12.5)} = \frac{3\ln 2 + \ln 5}{2\ln 5 - \ln 2}$$
Answer: $x = \frac{\ln(40)}{\ln(12.5)}$ or $\frac{3\ln 2 + \ln 5}{2\ln 5 - \ln 2}$.
Evaluate each logarithm without a calculator:
(a) $\log_3(243)$ (b) $\log_5\left(\frac{1}{25}\right)$ (c) $\log_7(\sqrt{7})$
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(a) $3^5 = 243 \implies \log_3(243) = \mathbf{5}$ [A1]
(b) $5^{-2} = \frac{1}{25} \implies \log_5\left(\frac{1}{25}\right) = \mathbf{-2}$ [A1]
(c) $7^{1/2} = \sqrt{7} \implies \log_7(\sqrt{7}) = \mathbf{\frac{1}{2}}$ [A1]
Express each in terms of simpler logarithms using log laws:
(a) $\log_b(x^3 y^2)$ (b) $\log_b\left(\frac{\sqrt{x}}{y^4}\right)$ (c) $\log_b\left(\frac{1}{x^5}\right)$
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(a) $\log_b(x^3) + \log_b(y^2) = 3\log_b(x) + 2\log_b(y)$ [A1]
(b) $\log_b(x^{1/2}) - \log_b(y^4) = \frac{1}{2}\log_b(x) - 4\log_b(y)$ [A1]
(c) $\log_b(x^{-5}) = -5\log_b(x)$ [A1]
Combine into a single logarithm with coefficient $1$: $$3\log_{10}(2) + \log_{10}(25) - \log_{10}(2)$$ Hence, find its exact integer value without a calculator.
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Step 1: Simplify like terms or apply power law [2 marks]
$3\log_{10}(2) - \log_{10}(2) = 2\log_{10}(2)$ [M1]
$2\log_{10}(2) + \log_{10}(25) = \log_{10}(2^2) + \log_{10}(25) = \log_{10}(4) + \log_{10}(25)$ [M1]
Step 2: Apply product law and evaluate [2 marks]
$\log_{10}(4 \times 25) = \log_{10}(100)$ [A1]
Since $10^2 = 100$, $\log_{10}(100) = \mathbf{2}$ [A1]
Solve for $x$: $$\log_2(x) + \log_2(x - 6) = 4$$
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Step 1: Apply product law [1 mark]
$\log_2(x(x - 6)) = 4$ [M1]
Step 2: Convert to exponential form [1 mark]
$x(x - 6) = 2^4 = 16 \implies x^2 - 6x - 16 = 0$ [M1]
Step 3: Solve and check domain [2 marks]
$(x - 8)(x + 2) = 0 \implies x = 8 \text{ or } x = -2$ [A1]
Domain requires $x > 0$ and $x - 6 > 0 \implies x > 6$. Reject $x = -2$.
Valid solution: $\mathbf{x = 8}$. [R1]
Solve $5^x = 42$ for $x$. Give your answer:
(a) In exact form using natural logarithms. [1 mark]
(b) Correct to 3 significant figures. [2 marks]
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(a) Exact form: [1 mark]
$\ln(5^x) = \ln(42) \implies x\ln 5 = \ln 42 \implies x = \frac{\ln 42}{\ln 5}$ [A1]
(b) Decimal to 3 s.f.: [2 marks]
$x = \frac{3.73767}{1.60944} \approx 2.3223\dots \approx \mathbf{2.32}$ [A2]
Prove the Change of Base Formula: $$\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$$ where $a, b, c > 0$ and $b, c \ne 1$. Show each step of logical reasoning clearly.
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Step 1: Define a variable [1 mark]
Let $x = \log_b(a)$ [M1].
Step 2: Convert to exponential form [1 mark]
By definition of logarithms: $b^x = a$ [A1].
Step 3: Take logarithm to new base $c$ [1 mark]
$\log_c(b^x) = \log_c(a)$
By the Power Law: $x \log_c(b) = \log_c(a)$ [M1].
Step 4: Solve for $x$ [1 mark]
Since $b \ne 1$, $\log_c(b) \ne 0$. Divide by $\log_c(b)$:
$x = \frac{\log_c(a)}{\log_c(b)} \implies \log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ $\blacksquare$ [R1]
Solve the quadratic equation in logarithmic form: $$(\log_3 x)^2 - 4\log_3 x - 12 = 0$$
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Step 1: Quadratic substitution [1 mark]
Let $u = \log_3 x$: $u^2 - 4u - 12 = 0$ [M1]
Step 2: Solve for $u$ [1 mark]
$(u - 6)(u + 2) = 0 \implies u = 6 \text{ or } u = -2$ [A1]
Step 3: Solve for $x$ [2 marks]
• $\log_3 x = 6 \implies x = 3^6 = \mathbf{729}$ [A1]
• $\log_3 x = -2 \implies x = 3^{-2} = \mathbf{\frac{1}{9}}$ [A1]
Both solutions satisfy $x > 0$. Solutions: $x = 729, \frac{1}{9}$.
In chemistry, the acidity of a liquid is measured by its $\text{pH}$ value, defined by the formula:
$$\text{pH} = -\log_{10}[\text{H}^+]$$
where $[\text{H}^+]$ is the hydrogen ion concentration in moles per litre ($\text{mol L}^{-1}$).
(a) Black coffee has $[\text{H}^+] = 1.0 \times 10^{-5}\text{ mol L}^{-1}$. Calculate its $\text{pH}$. [1 mark]
(b) Lemon juice has a $\text{pH}$ of $2.3$. Calculate its hydrogen ion concentration $[\text{H}^+]$ in scientific notation to 3 significant figures. [2 marks]
(c) How many times more concentrated in $[\text{H}^+]$ is lemon juice compared to coffee? [1 mark]
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(a) pH of coffee: [1 mark]
$\text{pH} = -\log_{10}(10^{-5}) = -(-5) = \mathbf{5.0}$ [A1]
(b) Concentration of lemon juice: [2 marks]
$-\log_{10}[\text{H}^+] = 2.3 \implies \log_{10}[\text{H}^+] = -2.3$ [M1]
$[\text{H}^+] = 10^{-2.3} \approx 5.0118 \times 10^{-3} \approx \mathbf{5.01 \times 10^{-3}\text{ mol L}^{-1}}$ [A1]
(c) Ratio: [1 mark]
$\text{Ratio} = \frac{10^{-2.3}}{10^{-5}} = 10^{5 - 2.3} = 10^{2.7} \approx \mathbf{501\text{ times}}$ [A1]
The Richter magnitude $M$ of an earthquake is given by $M = \log_{10}\left(\frac{I}{I_0}\right)$, where $I$ is the seismic wave intensity and $I_0$ is a baseline threshold.
(a) Earthquake A has magnitude $M_A = 7.8$ and Earthquake B has magnitude $M_B = 5.2$.
Show that $\frac{I_A}{I_B} = 10^{M_A - M_B}$. [2 marks]
(b) Calculate how many times more intense Earthquake A is than Earthquake B to the nearest integer. [2 marks]
(c) What increase in magnitude represents a 1000-fold increase in seismic intensity? [1 mark]
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(a) Proof of ratio: [2 marks]
$M_A = \log_{10}(I_A / I_0) \implies I_A = I_0 \cdot 10^{M_A}$
$M_B = \log_{10}(I_B / I_0) \implies I_B = I_0 \cdot 10^{M_B}$
$\frac{I_A}{I_B} = \frac{I_0 \cdot 10^{M_A}}{I_0 \cdot 10^{M_B}} = 10^{M_A - M_B}$ $\blacksquare$ [A2]
(b) Numerical comparison: [2 marks]
$\frac{I_A}{I_B} = 10^{7.8 - 5.2} = 10^{2.6} \approx 398.1 \approx \mathbf{398\text{ times}}$ [A2]
(c) 1000-fold increase: [1 mark]
$10^{\Delta M} = 1000 = 10^3 \implies \Delta M = \mathbf{3\text{ magnitude units}}$ [A1]
An investment of $\$5000$ earns $6.5\%$ annual compound interest compounded continuously, modeled by $A(t) = 5000 \cdot e^{0.065t}$, where $t$ is the number of years.
(a) Write an equation for the time $T_d$ required for the investment to double. [1 mark]
(b) Solve for $T_d$ analytically using natural logarithms. [2 marks]
(c) Calculate $T_d$ in years to 3 significant figures. [2 marks]
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(a) Doubling equation: [1 mark]
$5000 \cdot e^{0.065 T_d} = 10000 \implies e^{0.065 T_d} = 2$ [A1]
(b) Natural log solution: [2 marks]
Take $\ln$ of both sides: $\ln(e^{0.065 T_d}) = \ln 2$
$0.065 T_d = \ln 2 \implies T_d = \frac{\ln 2}{0.065}$ [A2]
(c) Value: [2 marks]
$T_d = \frac{0.693147}{0.065} \approx 10.663\dots \approx \mathbf{10.7\text{ years}}$ [A2]
Logarithms Quick Reference
Definition & Inversion
Operational Log Laws
Change of Base Formula
Scientific Log Models
Top 4 Extended Logarithm Pitfalls
1. Inventing Fake Distributive Laws
$\log(x + y) \ne \log x + \log y$! The logarithm of a sum CANNOT be split. Logarithms only split products: $\log(xy) = \log x + \log y$.
2. Confusing $\frac{\log x}{\log y}$ with $\log\left(\frac{x}{y}\right)$
$\log\left(\frac{x}{y}\right) = \log x - \log y$. In contrast, $\frac{\log x}{\log y} = \log_y x$ by change of base. They are completely different operations!
3. Forgetting Log Arguments Must Be Positive
$\log_b(x)$ is undefined for $x \le 0$. When solving log equations, always check that candidate solutions produce strictly positive arguments in all original logs.
4. Dropping Parentheses on Multi-Term Exponents
In $2^{x+1} = 5$, taking logs gives $(x + 1)\ln 2 = \ln 5 \implies x\ln 2 + \ln 2 = \ln 5$. Writing $x + 1 \ln 2$ without brackets leads to algebraic errors.
Logarithms Knowledge Check
Answer these 5 questions to test log definitions, operational rules, and exponential equations.
What is the exact value of $\log_2\left(\frac{1}{32}\right)$?
Which expression is equivalent to $\log_a(x) - 2\log_a(y) + 3\log_a(z)$?
Using the Change of Base Formula, which expression evaluates $\log_5(18)$?
Solve for $x$: $e^{2x} = 9$.
If the sound intensity $I$ increases by a factor of $100$, by how many decibels ($\text{dB}$) does the sound level $\beta = 10\log_{10}(I/I_0)$ increase?