Logarithms -- Diagnostic Tests

Tests edge cases, boundary conditions, and common misconceptions for logarithms.

Question:

Which of the following statements are true? Justify your answer.

$$\text{(i)}\; \log(ab) = \log a + \log b \qquad \text{(ii)}\; \log(a + b) = \log a + \log b$$

Solution:

(i) True. This is the product law of logarithms: $\log(ab) = \log a + \log b$.

(ii) False . Counterexample: let $a = b = 1$ with base 10.

$\log(1 + 1) = \log 2 \approx 0.301$.

$\log 1 + \log 1 = 0 + 0 = 0$.

$0.301 \neq 0$.

The logarithm function does NOT distribute over addition. $\log(a + b) \neq \log a + \log b$.

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Question:

Solve $\log_3(x - 2) + \log_3(x + 1) = 1$.

Solution:

Domain: $x - 2 > 0$ and $x + 1 > 0$ So $x > 2$.

Using the product law:

$$\log_3[(x-2)(x+1)] = 1$$ $$(x-2)(x+1) = 3$$ $$x^2 - x - 2 = 3$$ $$x^2 - x - 5 = 0$$ $$x = \frac{1 \pm \sqrt{1 + 20}}{2} = \frac{1 \pm \sqrt{21}}{2}$$

$x = \dfrac{1 + \sqrt{21}}{2} \approx 2.79$ (valid since $> 2$).

$x = \dfrac{1 - \sqrt{21}}{2} \approx -1.79$ (rejected: not in domain).

Solution: $x = \dfrac{1 + \sqrt{21}}{2}$.

A common mistake is forgetting to check the domain and accepting both roots.

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Question:

Given that $\log_2 3 = a$ and $\log_2 5 = b$ Express $\log_{15} 4$ in terms of $a$ and $b$.

Solution:

$$\log_{15} 4 = \frac{\log_2 4}{\log_2 15} = \frac{2}{\log_2(3 \times 5)} = \frac{2}{\log_2 3 + \log_2 5} = \frac{2}{a + b}$$

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Question:

Solve $(\log_2 x)^2 - 3\log_2 x - 4 = 0$.

Solution:

Let $u = \log_2 x$ (domain: $x > 0$).

$$u^2 - 3u - 4 = 0$$ $$(u - 4)(u + 1) = 0$$ $$u = 4 \quad \text{or} \quad u = -1$$

$\log_2 x = 4 \implies x = 16$.

$\log_2 x = -1 \implies x = \dfrac{1}{2}$.

Both satisfy $x > 0$.

Solution: $x = 16$ or $x = \dfrac{1}{2}$.

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Question:

Solve $2\log_x 4 - \log_4 x = 1$.

Solution:

Using change of base: $\log_x 4 = \dfrac{\log_4 4}{\log_4 x} = \dfrac{1}{\log_4 x}$.

Let $u = \log_4 x$ (domain: $x > 0$, $x \neq 1$ So $u \neq 0$).

$$\frac{2}{u} - u = 1$$ $$2 - u^2 = u$$ $$u^2 + u - 2 = 0$$ $$(u + 2)(u - 1) = 0$$

$u = -2$ or $u = 1$.

$u = -2$: $\log_4 x = -2 \implies x = 4^{-2} = \dfrac{1}{16}$.

$u = 1$: $\log_4 x = 1 \implies x = 4$.

Solution: $x = \dfrac{1}{16}$ or $x = 4$.

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Tests synthesis of logarithms with other topics.

Question:

Solve $3^{2x} - 4 \cdot 3^x + 3 = 0$.

Solution:

Let $u = 3^x$ ($u > 0$).

$$u^2 - 4u + 3 = 0$$ $$(u - 1)(u - 3) = 0$$

$u = 1$ or $u = 3$.

$3^x = 1 \implies x = 0$.

$3^x = 3 \implies x = 1$.

Solution: $x = 0$ or $x = 1$.

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Question:

Solve $\log_3(x + 4) > \log_3(8 - x)$.

Solution:

Since $\log_3$ is increasing, we can compare arguments directly:

$$x + 4 > 8 - x$$ $$2x > 4 \implies x > 2$$

Domain: $x + 4 > 0$ and $8 - x > 0$ Giving $-4 < x < 8$.

Combining: $2 < x < 8$.

Solution: $x \in (2,\; 8)$.

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Question:

Let $f(x) = \log_3(2x - 1)$. Find $f^{-1}(x)$ and its domain and range.

Solution:

$y = \log_3(2x - 1) \implies 3^y = 2x - 1 \implies x = \dfrac{3^y + 1}{2}$.

$f^{-1}(x) = \dfrac{3^x + 1}{2}$.

Domain of $f$: $2x - 1 > 0 \implies x > \dfrac{1}{2}$.

Range of $f$: $\mathbb{R}$ (since $3^y > 0$).

Domain of $f^{-1}$ = Range of $f$ = $\mathbb{R}$.

Range of $f^{-1}$ = Domain of $f$ = $\left(\dfrac{1}{2},\; \infty\right)$.

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Question:

Solve $\log_2(x + 4) + \log_2(x - 2) = 3$.

Solution:

Domain: $x + 4 > 0$ and $x - 2 > 0$ So $x > 2$.

Using the product law:

$$\log_2[(x+4)(x-2)] = 3$$ $$(x+4)(x-2) = 2^3 = 8$$ $$x^2 + 2x - 8 = 8$$ $$x^2 + 2x - 16 = 0$$ $$x = \frac{-2 \pm \sqrt{4 + 64}}{2} = \frac{-2 \pm \sqrt{68}}{2} = -1 \pm \sqrt{17}$$

$x = -1 + \sqrt{17} \approx 3.12$ (valid, $> 2$).

$x = -1 - \sqrt{17} \approx -5.12$ (rejected, not in domain).

Solution: $x = -1 + \sqrt{17}$.

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Question:

A population of bacteria doubles every 3 hours. If the initial population is 500, find:

(a) The population after 12 hours. (b) The time for the population to reach 32000.

Solution:

$P(t) = P_0 \cdot 2^{t/3}$ where $P_0 = 500$.

(a) $P(12) = 500 \cdot 2^{12/3} = 500 \cdot 2^4 = 500 \times 16 = 8000$.

(b) $500 \cdot 2^{t/3} = 32000$.

$2^{t/3} = 64 = 2^6$.

$\dfrac{t}{3} = 6 \implies t = 18$ hours.

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Question:

An earthquake has a magnitude of 7.2 on the Richter scale. How many times more powerful is it than an earthquake of magnitude 5.2?

Solution:

The Richter scale is logarithmic: each increase of 1 unit represents a 10-fold increase in amplitude.

Difference in magnitude: $7.2 - 5.2 = 2$.

The earthquake is $10^2 = 100$ times more powerful.

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Question:

Given that $\log_3 2 = a$ Express $\log_3 16 - \log_3 4$ in terms of $a$.

Solution:

$$\log_3 16 = \log_3 2^4 = 4\log_3 2 = 4a$$ $$\log_3 4 = \log_3 2^2 = 2\log_3 2 = 2a$$ $$\log_3 16 - \log_3 4 = 4a - 2a = 2a$$

Alternatively: $\log_3 16 - \log_3 4 = \log_3 \dfrac{16}{4} = \log_3 4 = 2a$.

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Question:

Solve $\log_{0.5}(2x - 1) > \log_{0.5}(x + 2)$.

Solution:

Domain: $2x - 1 > 0$ and $x + 2 > 0$ Giving $x > \dfrac{1}{2}$ and $x > -2$ So $x > \dfrac{1}{2}$.

Since the base $0.5$ is between 0 and 1, the logarithmic function is decreasing. Therefore the inequality reverses:

$$2x - 1 < x + 2$$ $$x < 3$$

Combining with the domain: $\dfrac{1}{2} < x < 3$.

Solution: $x \in \left(\dfrac{1}{2},\; 3\right)$.

DSE Exam Technique: Always check whether the logarithmic base is greater than or less than 1. This determines whether the function is increasing or decreasing, which affects the inequality direction. This is a frequent trap in DSE Paper 2.

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Question:

Given $\log_4 5 = a$ Find $\log_8 10$ in terms of $a$.

Solution:

$$\log_8 10 = \frac{\log_4 10}{\log_4 8}$$

$\log_4 10 = \log_4(2 \times 5) = \log_4 2 + \log_4 5 = \dfrac{1}{2} + a$.

$\log_4 8 = \log_4(4 \times 2) = 1 + \dfrac{1}{2} = \dfrac{3}{2}$.

$$\log_8 10 = \frac{\frac{1}{2} + a}{\frac{3}{2}} = \frac{1 + 2a}{3}$$

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Question:

Solve the simultaneous equations:

$\log_2 x + \log_2 y = 5$ $\log_2(x - y) = 1$

Solution:

From equation 1: $\log_2(xy) = 5 \implies xy = 32$. … (1)

From equation 2: $x - y = 2$. … (2)

From (2): $x = y + 2$. Substituting into (1):

$(y + 2)y = 32 \implies y^2 + 2y - 32 = 0$.

$$y = \frac{-2 \pm \sqrt{4 + 128}}{2} = \frac{-2 \pm \sqrt{132}}{2} = -1 \pm \sqrt{33}$$

Since $y > 0$: $y = -1 + \sqrt{33}$.

$x = y + 2 = 1 + \sqrt{33}$.

Check domain: $x > 0$, $y > 0$, $x - y > 0$. All satisfied.

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Question:

Solve $5^{x+1} = 2^{2x-1}$.

Solution:

Take logarithms of both sides (any base, say base 10 or natural log):

$$(x+1)\ln 5 = (2x - 1)\ln 2$$ $$x\ln 5 + \ln 5 = 2x\ln 2 - \ln 2$$ $$x\ln 5 - 2x\ln 2 = -\ln 2 - \ln 5$$ $$x(\ln 5 - 2\ln 2) = -(\ln 2 + \ln 5)$$ $$x = \frac{-(\ln 2 + \ln 5)}{\ln 5 - 2\ln 2} = \frac{-(\ln 10)}{\ln 5 - \ln 4} = \frac{-\ln 10}{\ln(5/4)} = \frac{\ln 10}{\ln(4/5)}$$ $$x = \frac{\ln 10}{\ln(4/5)} \approx \frac{2.303}{-0.223} \approx -10.33$$

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Asking “how many times?”: A logarithm asks “how many times must I multiply this base to get that number?”, it’s the inverse of exponentiation, like asking “what power gives me this result?”

Why it matters: Logarithms compress enormous ranges, earthquakes, pH, sound intensity, into manageable scales. They turn multiplication into addition, making complex calculations simpler.

The key insight: log(a) + log(b) = log(ab), not log(a+b), logarithms turn multiplication into addition, not addition into something else.

1. Forgetting domain restrictions in logarithmic equations. The argument of a logarithm must be strictly positive. After solving a logarithmic equation, always check that each solution satisfies the domain conditions. Solutions that make any log argument zero or negative must be rejected.

2. Not reversing the inequality for bases between 0 and 1. If $\log_b A > \log_b B$ and $0 < b < 1$ Then $A < B$ (not $A > B$). This is because the log function is decreasing when the base is between 0 and 1.

3. Incorrectly applying the power law. $\log(a^b) = b \log a$ But $(\log a)^b \neq b \log a$. The power law applies to the argument, not to the logarithm itself.

4. Assuming $\log a + \log b = \log(a + b)$. This is false. The correct law is $\log a + \log b = \log(ab)$. This is one of the most common algebraic errors in logarithms.

5. Using the wrong base in change of base formula. The change of base formula is $\log_a b = \dfrac{\log_c b}{\log_c a}$ for any valid base $c$. Swapping the numerator and denominator is a frequent error.

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(a) Solve $\log_3(x^2 - 3x + 2) = 1$. (3 marks) (b) Solve $\log_3(x^2 - 3x + 2) = 0$. (2 marks) (c) Hence solve $\log_3(x^2 - 3x + 2) \leq 1$. (3 marks)

Solution:

(a) $x^2 - 3x + 2 = 3^1 = 3$.

$x^2 - 3x - 1 = 0$.

$x = \dfrac{3 \pm \sqrt{9 + 4}}{2} = \dfrac{3 \pm \sqrt{13}}{2}$.

Domain check: $x^2 - 3x + 2 > 0 \implies (x-1)(x-2) > 0 \implies x < 1$ or $x > 2$.

$\dfrac{3 + \sqrt{13}}{2} \approx 3.30 > 2$ (valid). $\dfrac{3 - \sqrt{13}}{2} \approx -0.30 < 1$ (valid).

(b) $x^2 - 3x + 2 = 3^0 = 1$.

$x^2 - 3x + 1 = 0$.

$x = \dfrac{3 \pm \sqrt{5}}{2}$.

Domain check: both values satisfy the domain. $\dfrac{3+\sqrt{5}}{2} \approx 2.62 > 2$ and $\dfrac{3-\sqrt{5}}{2} \approx 0.38 < 1$.

(c) $\log_3(x^2 - 3x + 2) \leq 1 \implies 0 < x^2 - 3x + 2 \leq 3$.

$x^2 - 3x + 2 > 0$: $x < 1$ or $x > 2$.

$x^2 - 3x + 2 \leq 3 \implies x^2 - 3x - 1 \leq 0$: $\dfrac{3-\sqrt{13}}{2} \leq x \leq \dfrac{3+\sqrt{13}}{2}$.

Combining: $\dfrac{3-\sqrt{13}}{2} \leq x < 1$ or $2 < x \leq \dfrac{3+\sqrt{13}}{2}$.

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Let $f(x) = 2\log_3(x - 1) - \log_3(x^2 - 1)$.

(a) Find the domain of $f$. (2 marks) (b) Simplify $f(x)$. (3 marks) (c) Solve $f(x) = 1$. (3 marks)

Solution:

(a) $x - 1 > 0$ and $x^2 - 1 > 0$: $x > 1$.

(b) $f(x) = \log_3(x - 1)^2 - \log_3(x^2 - 1) = \log_3 \dfrac{(x-1)^2}{(x-1)(x+1)} = \log_3 \dfrac{x - 1}{x + 1}$ (for $x > 1$).

(c) $\log_3 \dfrac{x - 1}{x + 1} = 1 \implies \dfrac{x - 1}{x + 1} = 3 \implies x - 1 = 3x + 3 \implies -2x = 4 \implies x = -2$.

But $x > 1$ So $x = -2$ is rejected. No solution.

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The number of bacteria in a culture is given by $N = 1000 \times 2^{0.1t}$ Where $t$ is the time in hours.

(a) Find the initial number of bacteria. (1 mark) (b) Find the number of bacteria after 10 hours, giving your answer in exact form. (2 marks) (c) Find the time when the number of bacteria first exceeds 50000. (3 marks)

Solution:

(a) $t = 0$: $N = 1000 \times 2^0 = 1000$.

(b) $t = 10$: $N = 1000 \times 2^1 = 2000$.

(c) $1000 \times 2^{0.1t} > 50000$.

$2^{0.1t} > 50$.

$0.1t \log 2 > \log 50$.

$t > \dfrac{\log 50}{0.1 \log 2} = \dfrac{10 \log 50}{\log 2} = \dfrac{10(\log 100 - \log 2)}{\log 2} = \dfrac{10(2 - \log 2)}{\log 2} \approx \dfrac{10 \times 1.699}{0.301} \approx 56.4$ hours.

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If $\log_a 2 = p$ and $\log_a 5 = q$ Express the following in terms of $p$ and $q$:

(a) $\log_a 10$ (1 mark) (b) $\log_a 0.04$ (2 marks) (c) $\log_a 250$ (2 marks) (d) $\log_2 a$ (2 marks)

Solution:

(a) $\log_a 10 = \log_a(2 \times 5) = \log_a 2 + \log_a 5 = p + q$.

(b) $\log_a 0.04 = \log_a\left(\dfrac{4}{100}\right) = \log_a 4 - \log_a 100 = 2\log_a 2 - 2\log_a 10 = 2p - 2(p + q) = -2q$.

(c) $\log_a 250 = \log_a(125 \times 2) = \log_a 125 + \log_a 2 = 3\log_a 5 + p = 3q + p$.

(d) $\log_2 a = \dfrac{1}{\log_a 2} = \dfrac{1}{p}$.

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Solve the inequality $\log_2(x + 3) + \log_2(x - 1) \leq 3$.

Solution:

Domain: $x + 3 > 0$ and $x - 1 > 0$ So $x > 1$.

$$\log_2[(x+3)(x-1)] \leq 3$$ $$(x+3)(x-1) \leq 8$$ $$x^2 + 2x - 3 \leq 8$$ $$x^2 + 2x - 11 \leq 0$$ $$x = \frac{-2 \pm \sqrt{4 + 44}}{2} = \frac{-2 \pm \sqrt{48}}{2} = -1 \pm 2\sqrt{3}$$

Solution of quadratic: $-1 - 2\sqrt{3} \leq x \leq -1 + 2\sqrt{3}$.

Since $-1 + 2\sqrt{3} \approx 2.46$ and we need $x > 1$:

$1 < x \leq -1 + 2\sqrt{3}$ I.e. $x \in (1,\; -1 + 2\sqrt{3}]$.

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