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1
Find the value of \(\log_2 40 - \log_2 5 + \log_2 \dfrac{1}{4}\).
- A\(1\)
- B\(2\)
- C\(-1\)
- D\(5\)
- E\(3\)
- F\(0\)
Worked solution
\(\log_2\left(\frac{40}{5}\cdot\frac14\right) = \log_2 2 = 1\).
Full method — how and why
Combine everything into a single logarithm using the laws, then evaluate.
- Combine the first two: \(\log_2 40 - \log_2 5 = \log_2\dfrac{40}{5} = \log_2 8\).A difference of logs (same base) is the log of the quotient.
- Add the third: \(\log_2 8 + \log_2\frac14 = \log_2\left(8 \times \frac14\right) = \log_2 2\).A sum of logs is the log of the product. Note \(\log_2\frac14\) is negative, so this step reduces the total.
- Evaluate: \(\log_2 2 = 1\).\(\log_a a = 1\) for any valid base.
Where it goes wrong
- Treating \(\log_2\frac14\) as \(+2\) rather than \(-2\) and answering \(5\).
- Dividing the logarithms instead of subtracting them.
Takeaway \(\log\) turns \(\times\) into \(+\) and \(\div\) into \(-\); combine into one log before evaluating.
Answer A · Exponentials and logarithms (MM5.2) · question LOG-01
2
Solve \(3^{2x} = 4\). Give \(x\) in terms of logarithms to base \(3\).
- A\(x = \log_3 4 - 2\)
- B\(x = \log_3 2\)
- C\(x = \dfrac{\log_3 4}{4}\)
- D\(x = \log_2 3\)
- E\(x = \dfrac{1}{2}\log_3 2\)
- F\(x = 2\log_3 2\)
Worked solution
\(2x = \log_3 4 = 2\log_3 2\), so \(x = \log_3 2\).
Full method — how and why
Solving \(a^{kx} = b\). Take logs to the given base and simplify the right-hand side.
- Take \(\log_3\) of both sides: \(\log_3 3^{2x} = \log_3 4\), so \(2x = \log_3 4\).\(\log_3 3^{u} = u\) - logs and exponentials with the same base undo each other.
- Rewrite the right-hand side: \(\log_3 4 = \log_3 2^2 = 2\log_3 2\).The power law \(\log a^k = k\log a\) brings the exponent down.
- Divide by \(2\): \(x = \log_3 2\).The factor \(2\) cancels exactly, which is why the answer is so clean.
Where it goes wrong
- Stopping at \(x = \frac12\log_3 4\) and then mis-simplifying to \(\frac12 \log_3 2\).
- Writing \(\log_3 4 = \log_3 2 \times \log_3 2\).
Takeaway \(\log a^k = k \log a\); rewriting numbers as powers often makes fractions cancel.
Answer B · Exponentials and logarithms (MM5.3) · question LOG-02
3
Solve \(2^{2x} - 3\cdot 2^{x+2} + 32 = 0\). Find the sum of the solutions.
- A\(6\)
- B\(12\)
- C\(4\)
- D\(3\)
- E\(8\)
- F\(5\)
Worked solution
Let \(y = 2^x\): \(y^2 - 12y + 32 = (y - 4)(y - 8) = 0\), so \(2^x = 4\) or \(8\): \(x = 2\) or \(3\). Sum \(5\).
Full method — how and why
A hidden quadratic in \(2^x\). The middle term needs rewriting before the substitution is visible.
- Rewrite the middle term: \(3 \cdot 2^{x+2} = 3 \cdot 2^2 \cdot 2^x = 12 \cdot 2^x\).\(2^{x+2} = 2^x \times 4\) - separating the constant part is essential before substituting.
- Substitute \(y = 2^x\), noting \(2^{2x} = y^2\): \(y^2 - 12y + 32 = 0\).\(2^{2x} = (2^x)^2\); that is what makes it quadratic.
- Factorise: \((y-4)(y-8) = 0\), so \(y = 4\) or \(y = 8\).Both are positive, so both give solutions for \(x\).
- Convert back: \(2^x = 4 \Rightarrow x = 2\); \(2^x = 8 \Rightarrow x = 3\). Sum \(= 5\).The question asks for the sum of the \(x\) values, not the \(y\) values.
Where it goes wrong
- Summing the \(y\) values (\(4 + 8 = 12\)) - which is offered.
- Reading \(2^{x+2}\) as \(2^x + 2\).
Takeaway \(a^{x+k} = a^k a^x\): split off the constant before substituting, then convert the roots back to \(x\).
Answer F · Exponentials and logarithms (MM5.3) · question LOG-03
4
Given that \(\log_a x = 3\) and \(\log_a y = 5\), find \(\log_a \left(\dfrac{x^2 \sqrt{y}}{a}\right)\).
- A\(8\)
- B\(\dfrac{15}{2}\)
- C\(\dfrac{13}{2}\)
- D\(\dfrac{17}{2}\)
- E\(10\)
- F\(\dfrac{11}{2}\)
Worked solution
\(2\log_a x + \frac12\log_a y - \log_a a = 6 + \frac52 - 1 = \frac{15}{2}\).
Full method — how and why
Break the expression apart with the log laws, then substitute the given values.
- Split the quotient and product: \(\log_a\dfrac{x^2\sqrt y}{a} = \log_a x^2 + \log_a \sqrt y - \log_a a\).Products become sums, quotients become differences.
- Bring the powers down: \(2\log_a x + \frac12\log_a y - \log_a a\).\(\sqrt y = y^{1/2}\), so its log is \(\frac12\log_a y\).
- Substitute \(\log_a x = 3\), \(\log_a y = 5\), \(\log_a a = 1\): \(6 + \frac52 - 1\).\(\log_a a = 1\) always - a fact worth having instantly available.
- Add: \(\frac{12 + 5 - 2}{2} = \frac{15}{2}\).Put everything over \(2\).
Where it goes wrong
- Forgetting \(\log_a a = 1\) and answering \(\frac{17}{2}\).
- Using \(\log_a\sqrt y = \sqrt{\log_a y}\).
Takeaway Expand fully with the three laws before substituting; \(\log_a a = 1\) and \(\log_a 1 = 0\).
Answer B · Exponentials and logarithms (MM5.2) · question LOG-04
5
Solve \(\log_3(x + 6) + \log_3(x - 2) = 2\).
- A\(x = 5\)
- B\(x = -7\) or \(x = 3\)
- C\(x = 3\)
- D\(x = 3\) or \(x = 7\)
- E\(x = -3\)
- F\(x = 1\)
Worked solution
\((x+6)(x-2) = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\). \(x = -7\) makes both logs undefined, so \(x = 3\) only.
Full method — how and why
Combine the logs, convert to an equation, then check the domain - that is what this question is really testing.
- Combine: \(\log_3\left[(x+6)(x-2)\right] = 2\).Sum of logs is the log of the product.
- Convert to exponential form: \((x+6)(x-2) = 3^2 = 9\).\(\log_a N = c\) means \(N = a^c\).
- Expand and solve: \(x^2 + 4x - 12 = 9 \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow (x+7)(x-3) = 0\), so \(x = -7\) or \(x = 3\).Ordinary quadratic work.
- Check both in the original equation: \(x = 3\) gives \(\log_3 9 + \log_3 1\) ✓; \(x = -7\) gives \(\log_3(-1)\), which is undefined. So \(x = 3\) only.Logarithms of negative numbers (or zero) do not exist, so roots that break the original expression must be rejected.
Where it goes wrong
- Giving both roots - the option '\(x = -7\) or \(x = 3\)' exists precisely to catch this.
- Checking only one of the two logarithms.
Takeaway Always substitute your solutions back: combining logs can create solutions the original equation never had.
Answer C · Exponentials and logarithms (MM5.2) · question LOG-05
6
Which of the following statements is/are true for all positive real \(x\) and \(y\)?
- I\(\log_{10}(x + y) = \log_{10} x + \log_{10} y\)
- II\(\log_{10}(xy)^2 = 2\log_{10} x + 2\log_{10} y\)
- III\(\log_{10} x - \log_{10} y = \log_{10} x \div \log_{10} y\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
II is correct: \(\log(xy)^2 = 2\log(xy) = 2\log x + 2\log y\). I and III confuse the laws: \(\log\) of a sum has no simple form, and a difference of logs is \(\log\) of a quotient, not a quotient of logs.
Full method — how and why
Three log 'laws', two of which are fabrications. Test each against the genuine rules.
- I: \(\log(x+y)\) has no expansion. Test \(x = y = 1\): \(\log 2 \approx 0.30\) but \(\log 1 + \log 1 = 0\). False.The product law is \(\log(xy) = \log x + \log y\); the sum inside is a different animal entirely.
- II: \(\log(xy)^2 = 2\log(xy) = 2(\log x + \log y) = 2\log x + 2\log y\). True.Power law first, then product law - both applied correctly.
- III: \(\log x - \log y = \log\frac xy\), which is not the same as \(\frac{\log x}{\log y}\). Test \(x = 100\), \(y = 10\): left \(= 1\), right \(= \frac21 = 2\). False.Difference of logs is the log of a quotient, not the quotient of logs.
- II only.Match to the options.
Where it goes wrong
- Accepting I by analogy with the product law.
- Accepting III because 'subtraction corresponds to division' - it does, but inside the log.
Takeaway The laws act on what is inside the log: \(\log(xy)\), \(\log\frac xy\), \(\log x^k\). Nothing simplifies \(\log(x+y)\).
Answer C · Exponentials and logarithms (MM5.2) · question LOG-07
7
Find the value of \(\dfrac{\log_5 27}{\log_5 3}\).
- A\(9\)
- B\(\dfrac{27}{3}\)
- C\(\log_5 9\)
- D\(3\)
- E\(\log_5 24\)
- F\(24\)
Worked solution
\(\log_5 27 = \log_5 3^3 = 3\log_5 3\), so the quotient is \(3\).
Full method — how and why
A quotient of two logs with the same base. Rewriting the numerator's argument as a power makes it cancel.
- Write \(27 = 3^3\), so \(\log_5 27 = \log_5 3^3 = 3\log_5 3\).The power law applies to the argument, whatever the base of the log.
- Form the quotient: \(\dfrac{3\log_5 3}{\log_5 3} = 3\).The common factor \(\log_5 3\) cancels (it is non-zero).
Where it goes wrong
- Subtracting to get \(\log_5 24\) - a quotient of logs is not the log of a quotient.
- Dividing the arguments to get \(\frac{27}{3} = 9\) in log form.
Takeaway \(\frac{\log_c a}{\log_c b} = \log_b a\): quotients of logs change the base, they do not divide the arguments.
Answer D · Exponentials and logarithms (MM5.2) · question LOG-09
8
How many real solutions does the equation \(4^{x} + 2^{x+1} - 8 = 0\) have?
- A\(1\)
- B\(2\)
- C\(0\)
- D\(3\)
- Einfinitely many
Worked solution
\(y = 2^x\): \(y^2 + 2y - 8 = (y + 4)(y - 2) = 0\). \(y = -4\) is impossible, \(y = 2\) gives \(x = 1\). One solution.
Full method — how and why
Quadratic in \(2^x\) again, but this time one root must be rejected - which changes the count.
- Rewrite: \(4^x = (2^2)^x = (2^x)^2\) and \(2^{x+1} = 2\cdot 2^x\).Get everything in terms of \(2^x\) before substituting.
- Let \(y = 2^x\): \(y^2 + 2y - 8 = 0\), which factorises as \((y+4)(y-2) = 0\).Standard quadratic.
- Reject \(y = -4\): \(2^x\) is positive for every real \(x\), so it can never equal a negative number.This is the crucial filter - exponentials are always strictly positive.
- From \(y = 2\): \(2^x = 2\), so \(x = 1\). Exactly one real solution.The question asks how many, so the count is the answer.
Where it goes wrong
- Answering \(2\) because the quadratic has two roots.
- Rejecting \(y = 2\) as well, and answering \(0\).
Takeaway After substituting \(y = a^x\), discard every root with \(y \leq 0\) - then count what remains.
Answer A · Exponentials and logarithms (MM5.3) · question LOG-10
9
Solve \(\log_2 x + \log_4 x = 6\). (You may use \(\log_4 x = \dfrac{1}{2}\log_2 x\).)
- A\(x = 8\)
- B\(x = 2^{6}\)
- C\(x = 16\)
- D\(x = 64\)
- E\(x = 4\)
- F\(x = 32\)
Worked solution
\(\frac32\log_2 x = 6 \Rightarrow \log_2 x = 4 \Rightarrow x = 16\).
Full method — how and why
Two different bases. The hint converts one to the other, after which it is a linear equation in \(\log_2 x\).
- Use the given relation \(\log_4 x = \frac12\log_2 x\).It holds because \(4 = 2^2\): it takes twice as many factors of \(2\) as of \(4\) to build \(x\).
- Substitute: \(\log_2 x + \frac12\log_2 x = 6\), i.e. \(\frac32\log_2 x = 6\).Both terms now involve the same quantity, so treat \(\log_2 x\) as a single unknown.
- Solve for the log: \(\log_2 x = 4\).Multiply by \(\frac23\).
- Convert: \(x = 2^4 = 16\).\(\log_2 x = 4\) means \(x = 2^4\).
Where it goes wrong
- Solving \(\log_2 x = 6\) and answering \(64\) - which is offered (twice, as \(64\) and \(2^6\)).
- Adding the bases or the arguments instead of using the relation.
Takeaway Get every logarithm to the same base first; then it is ordinary algebra in that one quantity.
Answer C · Exponentials and logarithms (MM5.2) · question LOG-11
10
The curve \(y = a^{x}\) passes through \((2, 12)\) and \((4, 108)\). Find \(a\).
- A\(4\)
- B\(3\)
- C\(\sqrt{12}\)
- D\(9\)
- E\(2\)
- F\(6\)
Worked solution
\(\frac{a^4}{a^2} = a^2 = \frac{108}{12} = 9\), so \(a = 3\) (as \(a > 0\)).
Full method — how and why
Two points on an exponential curve. Dividing eliminates any constant and isolates a power of \(a\).
- Write both facts: \(a^2 = 12\) and \(a^4 = 108\).Substitute each point into \(y = a^x\).
- Divide: \(\dfrac{a^4}{a^2} = a^2 = \dfrac{108}{12} = 9\).Dividing subtracts indices, giving a single power of \(a\) - the standard move with two points.
- Solve: \(a = 3\) (bases of exponentials are positive).Reject \(a = -3\): \(y = a^x\) requires \(a > 0\).
Where it goes wrong
- Answering \(9\) - that is \(a^2\), not \(a\).
- Subtracting the \(y\)-values instead of dividing.
Takeaway Divide two points on an exponential curve; the index difference gives a clean power of the base.
Answer B · Exponentials and logarithms (MM5.1) · question LOG-12
11
Given that \(\log_{10} 2 = p\) and \(\log_{10} 3 = q\), express \(\log_{10} 0.75\) in terms of \(p\) and \(q\).
- A\(3q - 4p\)
- B\(\dfrac{q}{2p}\)
- C\(2p - q\)
- D\(q + 2p\)
- E\(q - p\)
- F\(q - 2p\)
Worked solution
\(0.75 = \frac34\), so \(\log 0.75 = \log 3 - \log 4 = q - 2p\).
Full method — how and why
Express the number in terms of the given primes, then apply the log laws.
- Write \(0.75 = \dfrac34\).Decimals must become fractions of small numbers before the laws help.
- Split: \(\log_{10}\frac34 = \log_{10}3 - \log_{10}4\).Quotient law.
- Write \(4 = 2^2\): \(\log_{10}4 = 2\log_{10}2 = 2p\).Only \(\log 2\) and \(\log 3\) are available, so every number must be built from \(2\)s and \(3\)s.
- Combine: \(q - 2p\).Substituting \(\log_{10}3 = q\).
Where it goes wrong
- Using \(\log 4 = p^2\) or \(= 2 + p\).
- Getting the subtraction backwards and answering \(2p - q\).
Takeaway Factorise the argument into the primes you have been given, then apply product/quotient/power laws.
Answer F · Exponentials and logarithms (MM5.2) · question LOG-13
12
Solve \(\log_2 (x^2 - 4) - \log_2 (x - 2) = 3\).
- A\(x = 2\)
- B\(x = 6\)
- C\(x = -10\) or \(x = 6\)
- D\(x = 8\)
- E\(x = 10\)
- F\(x = 4\)
Worked solution
For \(x > 2\): \(\log_2\frac{(x-2)(x+2)}{x-2} = \log_2(x + 2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).
Full method — how and why
Combine the logs; the algebraic fraction simplifies dramatically because the numerator factorises.
- Combine: \(\log_2\dfrac{x^2-4}{x-2} = 3\).Difference of logs is the log of the quotient.
- Factorise and cancel: \(\dfrac{(x-2)(x+2)}{x-2} = x + 2\) (valid since \(x \neq 2\)).The domain requires \(x > 2\) for \(\log_2(x-2)\) to exist, so cancelling is legitimate.
- Convert: \(\log_2(x+2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).\(2^3 = 8\).
- Check the domain: \(x = 6\) gives \(x^2 - 4 = 32 > 0\) and \(x - 2 = 4 > 0\) ✓.Always verify against the original.
Where it goes wrong
- Forgetting the domain and including a spurious negative solution.
- Cancelling incorrectly to get \(x - 2\) or \(x^2 - 2\).
Takeaway Factorise before cancelling inside a log, and confirm every solution keeps all arguments positive.
Answer B · Exponentials and logarithms (MM5.2) · question LOG-15
13
For which of the following values of \(x\) is \(3^{x} < 2^{x+1}\)?
- I\(x = 0\)
- II\(x = 1\)
- III\(x = 2\)
- Anone of them
- BI only
- CII only
- DIII only
- EI and II only
- FI and III only
- GII and III only
- HI, II and III
Worked solution
\(x = 0\): \(1 < 2\) true. \(x = 1\): \(3 < 4\) true. \(x = 2\): \(9 < 8\) false. So I and II.
Full method — how and why
Three numerical tests. Nothing conceptual - just careful arithmetic with small powers.
- I: \(x = 0\) gives \(3^0 = 1\) and \(2^{1} = 2\). Is \(1 < 2\)? Yes. True.\(a^0 = 1\) for any \(a\).
- II: \(x = 1\) gives \(3^1 = 3\) and \(2^2 = 4\). Is \(3 < 4\)? Yes. True.Careful with the index \(x + 1 = 2\).
- III: \(x = 2\) gives \(3^2 = 9\) and \(2^3 = 8\). Is \(9 < 8\)? No. False.\(3^x\) has overtaken \(2^{x+1}\) by here - exponential growth with the larger base always wins eventually.
- I and II only.Match to the option list.
Where it goes wrong
- Computing \(2^{x+1}\) as \(2^x + 1\).
- Assuming the inequality holds for all \(x\) because it holds for the first two.
Takeaway Substitute each value carefully; a larger base always overtakes eventually, whatever the head start.
Answer E · Exponentials and logarithms (MM5.1) · question LOG-16
14
If \(\log_a b = 2\) and \(\log_b c = 3\), find \(\log_a c\).
- A\(8\)
- B\(6\)
- C\(\dfrac{2}{3}\)
- D\(9\)
- E\(5\)
- F\(\dfrac{3}{2}\)
Worked solution
\(b = a^2\) and \(c = b^3 = a^6\), so \(\log_a c = 6\).
Full method — how and why
A chain of logarithms. Convert each statement to exponential form and substitute.
- \(\log_a b = 2\) means \(b = a^2\).\(\log_a b = c \Leftrightarrow a^c = b\).
- \(\log_b c = 3\) means \(c = b^3\).Same conversion with the next pair.
- Substitute: \(c = (a^2)^3 = a^6\).Powers of powers multiply.
- Therefore \(\log_a c = 6\).Read the exponent straight off. (In general \(\log_a c = \log_a b \times \log_b c = 2 \times 3\).)
Where it goes wrong
- Adding the values to get \(5\).
- Reading \((a^2)^3\) as \(a^5\).
Takeaway Logarithms chain by multiplication: \(\log_a c = \log_a b \times \log_b c\).
Answer B · Exponentials and logarithms (MM5.2) · question LOG-18
15
The graph of \(y = 2^{x}\) is translated \(3\) units in the positive \(x\)-direction. The resulting graph has equation
- A\(y = \dfrac{1}{8}\cdot 2^{x}\)
- B\(y = 2^{x/3}\)
- C\(y = 8\cdot 2^{x}\)
- D\(y = 2^{x} - 3\)
- E\(y = 2^{3x}\)
- F\(y = 2^{x} + 3\)
Worked solution
\(y = 2^{x - 3} = 2^{-3}\cdot 2^x = \frac18 \cdot 2^x\). (A translation of an exponential is also a stretch.)
Full method — how and why
A translation of an exponential graph is also a stretch. The algebra shows why.
- Translating \(3\) right replaces \(x\) by \(x - 3\): \(y = 2^{x-3}\).A shift to the right by \(a\) substitutes \(x - a\).
- Split the index: \(2^{x-3} = 2^x \times 2^{-3} = \dfrac{1}{8}\cdot 2^x\).\(a^{m-n} = \frac{a^m}{a^n}\) - the translation has become multiplication by a constant.
- So the new graph is the old one stretched vertically by factor \(\frac18\).This is special to exponentials: horizontal translation and vertical stretch are the same thing.
Where it goes wrong
- Writing \(2^{x-3} = 2^x - 3\) and choosing that option.
- Using \(2^{+3} = 8\) as the factor, which corresponds to a shift to the left.
Takeaway \(a^{x-k} = a^{-k}a^x\): for exponentials, translating horizontally is the same as stretching vertically.
Answer A · Exponentials and logarithms (MM5.1) · question LOG-06
16
Given that \(5^{x} = 2\) and \(2^{y} = 125\), find the value of \(xy\).
- A\(1\)
- B\(3\)
- C\(5\)
- D\(2\)
- E\(6\)
- F\(\log_5 2 \cdot \log_2 125\)
Worked solution
\(5^{xy} = (5^x)^y = 2^y = 125 = 5^3\), so \(xy = 3\).
Full method — how and why
Two exponential facts that chain together. Raising the first to the power \(y\) links them.
- Raise the first equation to the power \(y\): \((5^x)^y = 2^y\), i.e. \(5^{xy} = 2^y\).\((a^x)^y = a^{xy}\) - this is what connects the two statements.
- Substitute the second fact: \(2^y = 125\).Given directly.
- Write \(125\) as a power of \(5\): \(5^{xy} = 5^3\).Both sides are now powers of \(5\).
- Equate the indices: \(xy = 3\).Valid because \(5^u\) is one-to-one.
Where it goes wrong
- Trying to compute \(x\) and \(y\) separately as logarithms and getting stuck without a calculator.
- Multiplying the two equations to get \(10^{x+y} = 250\), which leads nowhere.
Takeaway To find a product of exponents, raise one equation to the other's exponent and chain the results.
Answer B · Exponentials and logarithms (MM5.3) · question LOG-08
17
The number of integers \(n\) satisfying \(2^{n} < 1000 < 2^{n+3}\) is
- A\(3\)
- B\(4\)
- C\(0\)
- D\(1\)
- E\(10\)
- F\(2\)
Worked solution
\(2^9 = 512 < 1000 < 1024 = 2^{10}\). Need \(n \leq 9\) and \(n + 3 \geq 10\), so \(n = 7, 8, 9\): three integers.
Full method — how and why
A double inequality in powers of \(2\). Locate \(1000\) between consecutive powers, then translate into conditions on \(n\).
- Find where \(1000\) sits: \(2^9 = 512\) and \(2^{10} = 1024\), so \(2^9 < 1000 < 2^{10}\).Powers of \(2\) up to \(1024\) are worth knowing by heart.
- The left condition \(2^n < 1000\) requires \(n \leq 9\).\(2^{10} = 1024 > 1000\), so \(n = 10\) fails.
- The right condition \(1000 < 2^{n+3}\) requires \(n + 3 \geq 10\), i.e. \(n \geq 7\).\(2^{9} = 512 < 1000\), so \(n + 3 = 9\) (i.e. \(n = 6\)) fails.
- Combine: \(7 \leq n \leq 9\), giving \(n = 7, 8, 9\) - three integers.Count the integers in the range, inclusive.
Where it goes wrong
- Counting \(9 - 7 = 2\) instead of \(3\).
- Using strict inequalities on \(n\) where the powers allow equality.
Takeaway Convert conditions on \(2^n\) into conditions on \(n\) by comparing with the nearest powers, then count inclusively.
Answer A · Exponentials and logarithms (MM5.3) · question LOG-14
18
The equation \(9^{x} - k\cdot 3^{x} + 9 = 0\) has exactly one real solution. Find the value of \(k\).
- A\(3\)
- B\(-6\)
- C\(\pm 6\)
- D\(9\)
- E\(6\)
- F\(0\)
Worked solution
With \(y = 3^x > 0\): \(y^2 - ky + 9 = 0\). Exactly one solution for \(x\) means exactly one positive root: a repeated root (\(k^2 = 36\), product \(9 > 0\) so the roots have the sign of \(k\)): \(k = 6\) (repeated root \(y = 3\), \(x = 1\)). \(k = -6\) gives \(y = -3\), no solutions. (Two distinct positive roots would give two solutions.)
Full method — how and why
A quadratic in \(3^x\) with a parameter. 'Exactly one real solution' is a condition on the positive roots, not simply on the discriminant.
- Substitute \(y = 3^x\), noting \(y > 0\): \(y^2 - ky + 9 = 0\).\(9^x = (3^x)^2\). The restriction \(y > 0\) is the heart of the question.
- The product of the roots is \(9 > 0\), so the roots have the same sign; their sum is \(k\), so both are positive exactly when \(k > 0\).Sum and product of roots decide the signs without solving.
- For exactly one value of \(x\), we need exactly one positive root, which (given both roots share a sign) means a repeated root: discriminant \(k^2 - 36 = 0\), so \(k = \pm6\).Two distinct positive roots would give two solutions for \(x\); two negative roots would give none.
- Check the signs: \(k = 6\) gives the repeated root \(y = 3 > 0\), so \(x = 1\) - exactly one solution. \(k = -6\) gives \(y = -3 < 0\), so no solutions at all. Hence \(k = 6\).The final sign check is what rules out \(-6\), which is offered.
Where it goes wrong
- Answering \(\pm6\) from the discriminant alone.
- Forgetting that \(y\) must be positive and counting \(y = -3\) as a solution.
Takeaway With \(y = a^x\), translate conditions on \(x\) into conditions on the positive roots in \(y\), using sum and product to check signs.
Answer E · Exponentials and logarithms (MM5.3) · question LOG-17
19
A quantity halves every \(5\) hours. Which expression gives the number of hours for it to fall to \(\dfrac{1}{10}\) of its initial value?
- A\(5\log_2 5\)
- B\(\dfrac{\log_{10} 2}{5}\)
- C\(5\log_{10} 2\)
- D\(\dfrac{5}{\log_{10} 2}\)
- E\(25\)
- F\(50\log_{10} 2\)
Worked solution
\(\left(\frac12\right)^{t/5} = \frac{1}{10}\). Taking logs to base \(10\): \(-\frac t5 \log_{10} 2 = -1\), so \(t = \dfrac{5}{\log_{10} 2}\) (about \(16.6\) hours).
Full method — how and why
Exponential decay. Build the model, then take logs - and be careful which expression the answer takes.
- Model: after \(t\) hours the quantity is multiplied by \(\left(\frac12\right)^{t/5}\), because \(t/5\) half-lives have passed.Dividing \(t\) by the half-life counts how many halvings have occurred.
- Set it equal to \(\frac{1}{10}\): \(\left(\frac12\right)^{t/5} = \dfrac{1}{10}\).'Falls to one tenth' is a statement about the multiplier.
- Take \(\log_{10}\): \(\dfrac t5\log_{10}\dfrac12 = \log_{10}\dfrac{1}{10} = -1\), i.e. \(-\dfrac t5\log_{10}2 = -1\).\(\log\frac12 = -\log 2\) and \(\log\frac1{10} = -1\); the two minus signs will cancel.
- Solve: \(t = \dfrac{5}{\log_{10}2}\) (about \(16.6\) hours).Multiply both sides by \(-\frac{5}{\log_{10}2}\).
Where it goes wrong
- Answering \(5\log_{10}2\) - the reciprocal, which would be under a second.
- Sanity check: the quantity halves in \(5\) h and \(\frac1{10} < \frac18\), so \(t\) must exceed \(15\) h; \(5\log_{10}2 \approx 1.5\) is obviously wrong.
Takeaway Set up the multiplier, take logs, and sanity-check the size of the answer against a couple of half-lives.
Answer D · Exponentials and logarithms (MM5.3) · question LOG-19
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Which of the following is a counterexample to the claim 'for all positive \(x\) and \(y\), \(\log_{10}(x + y) \leq \log_{10} x + \log_{10} y\)'?
- A\(x = 10\), \(y = 100\)
- B\(x = y = 2\)
- C\(x = 5\), \(y = 5\)
- D\(x = 4\), \(y = 4\)
- E\(x = y = 1\)
Worked solution
\(x = y = 1\): \(\log 2 > 0 = \log 1 + \log 1\), so the inequality fails. For the others \(x + y \leq xy\) and the claim holds (\(\log\) is increasing).
Full method — how and why
To break an inequality claim, look for the case where the two sides are closest - here small values, where logs go to zero or negative.
- Test \(x = y = 1\): left side \(\log_{10}2 \approx 0.30\); right side \(\log_{10}1 + \log_{10}1 = 0\).\(\log 1 = 0\) for any base, so the right side collapses.
- \(0.30 > 0\), so the claimed inequality \(\leq\) fails. Counterexample.Hypothesis satisfied (\(x,y\) positive), conclusion false.
- Why the others work: \(\log x + \log y = \log(xy)\), and \(\log\) is increasing, so the claim is just \(x+y \leq xy\). For \(x=y=2\): \(4 \leq 4\) ✓; \(x=y=4\): \(8 \leq 16\) ✓; \(x=y=5\): \(10\leq25\) ✓; \(x=10,y=100\): \(110 \leq 1000\) ✓.Converting the log inequality into an algebraic one makes all five cases quick to check.
Where it goes wrong
- Trying large values, where \(xy\) dwarfs \(x+y\) and the claim comfortably holds.
- Forgetting \(\log_{10}1 = 0\) and miscomputing the right side.
- Thinking \(\log(x+y) = \log x + \log y\) - that is the error the whole question is built on.
Takeaway \(\log x + \log y = \log(xy)\), never \(\log(x+y)\). Test inequality claims near the boundary, where the sides are closest.
Answer E · Exponentials and logarithms (MM5.2) · question X2-06
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Consider the statement: for all real \(x\), \(2^{x} > x^2\). Which of the following is a counterexample?
- A\(x = 5\)
- B\(x = 3\)
- C\(x = 0\)
- D\(x = 10\)
- E\(x = 1\)
Worked solution
\(2^3 = 8 < 9 = 3^2\), so \(x = 3\) breaks the claim. The other options satisfy it: \(2 > 1\), \(32 > 25\), \(1 > 0\), \(1024 > 100\).
Full method — how and why
Exponential versus quadratic growth: the exponential eventually wins, but there is a window where it loses. Test values inside that window.
- Check the candidates: \(x=0\) gives \(1>0\) ✓; \(x=1\) gives \(2>1\) ✓; \(x=5\) gives \(32>25\) ✓; \(x=10\) gives \(1024>100\) ✓.Evaluate both sides for each - no cleverness needed.
- \(x=3\): \(2^3 = 8\) and \(3^2 = 9\). Since \(8 < 9\), the claim fails. Counterexample.Hypothesis (\(x\) real) satisfied, conclusion false.
- Why here? \(2^x\) and \(x^2\) cross at \(x=2\) (both \(4\)) and again near \(x=4\) (both \(16\)), so the quadratic is ahead between them.Knowing where the curves cross tells you exactly where to hunt.
Where it goes wrong
- Testing only large \(x\), where the exponential dominates and the claim holds.
- Overlooking the window: the two sides are equal at \(x=2\) and \(x=4\), so the quadratic leads strictly between them - that is where a counterexample must lie.
- Assuming exponentials always beat polynomials - true only eventually.
Takeaway \(2^x\) and \(x^2\) meet at \(x=2\) and \(x=4\); between them the quadratic is larger. 'Eventually bigger' is not 'always bigger'.
Answer B · Exponentials and logarithms (MM5.3) · question X2-22
How this topic has been examined
Real TMUA questions on Exponentials and logarithms, newest first. Links open the official paper at the question and the online marking page.