Indices and Logarithms

14 Grade 10 Core Mathematics questions on Indices and Logarithms, with worked marking schemes. 14 questions are available on this topic in full.

1. Question 1

   3 marks

   Express as a single logarithm: 2 log x + log y - 3 log z.

   (3 marks)

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2. Question 2

   3 marks

   State, in words, what a negative index $a^{-n}$ means, and hence evaluate $(\frac{2}{5}{)}^{-2}$, giving your answer as a fraction in its simplest form.

   (3 marks)

   Marking scheme

   1 mark: $a^{-n}$ means the reciprocal of $a^n$, that is $a^{-n}=1/a^n$ (so a negative index inverts the base). 1 mark: $(\frac{2}{5}{)}^{-2}=(\frac{5}{2}{)}^2$. 1 mark: = $\frac{25}{4}$ (accept 6 $\frac{1}{4}$). (Accept the route $1÷(\frac{2}{5}{)}^2=1÷\frac{4}{25}$ for the second mark.)

3. Question 3

   1 mark

   Evaluate 27^(2/3).

   A. 9

   B. 6

   C. 18

   D. 54

   (1 mark)

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4. Question 4

   3 marks

   Solve for x: $3^{2x-1}=81$.

   (3 marks)

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5. Question 5

   1 mark

   The statement $4^3=64$ written in logarithmic form is:

   A. log₃ 64 = 4

   B. log₆₄ 4 = 3

   C. log₄ 64 = 3

   D. log₄ 3 = 64

   (1 mark)

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6. Question 6

   5 marks

   A dairy farmer in Nyandarua starts with 40 cows and the herd grows by 8% each year, so that after n years the herd is 40$(1.08{)}^n$. Using logarithms, determine the least number of whole years after which the herd will exceed 80 cows.

   (5 marks)

   Marking scheme

   1 mark: 40$(1.08{)}^n$ > 80, so $(1.08{)}^n$ > 2. 1 mark: taking logarithms of both sides, n log 1.08 > log 2. 1 mark: n > $\log 2/\log 1.08=\frac{0.3010}{0.0334}$. 1 mark: n > 9.01 (accept 9.0 to 9.1). 1 mark: least whole number of years = 10. (Award the final mark only for 10; an answer of 9 shows the inequality was not respected.)

7. Question 7

   1 mark

   Simplify $(3x^2{y}^3{)}^3$.

   A. 27$x^6$$y^9$

   B. 9$x^6$$y^9$

   C. 27$x^5$$y^6$

   D. 3$x^6$$y^9$

   (1 mark)

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8. Question 8

   1 mark

   Evaluate $5^0+2^{-2}$.

   A. 5.25

   B. 0.25

   C. 1.5

   D. 1.25

   (1 mark)

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9. Question 9

   1 mark

   Given that log 2 = 0.3010 and log 3 = 0.4771, find log 12.

   A. 1.4343

   B. 0.1433

   C. 1.0791

   D. 0.7781

   (1 mark)

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10. Question 10

    4 marks

    Given that log 5 = 0.6990, find, without using tables or a calculator, the value of log 500 and log 0.5.

    (4 marks)

    Marking scheme

    1 mark: $\log 500=\log (5\times 100)=\log 5+\log 100$. 1 mark: = 0.6990 + 2 = 2.6990. 1 mark: $\log 0.5=\log (5÷10)=\log 5-\log 10$. 1 mark: = 0.6990 - 1 = -0.3010 (accept 1̄.6990 in bar notation).

11. Question 11

    1 mark

    Solve for x: log x + log 4 = log 20, where all logarithms are to base 10.

    A. x = 16

    B. x = 5

    C. x = 80

    D. x = 24

    (1 mark)

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12. Question 12

    1 mark

    Simplify log 45 - log 9 + log 2, where all logarithms are to base 10.

    A. log 4

    B. log 810

    C. log 10

    D. log 38

    (1 mark)

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13. Question 13

    4 marks

    Outline two situations in everyday Kenyan life where numbers are conveniently written using indices or logarithms, and in each case explain briefly why that form is convenient.

    (4 marks)

    Marking scheme

    1 mark each for any two valid situations, plus 1 mark each for a matching explanation, to a maximum of 4. Acceptable situations include: (i) population or bank interest growth — compound growth is repeated multiplication, so an index shows the number of growth periods directly; (ii) earthquake magnitude on the Richter scale — a logarithm compresses a huge range of energies into a small readable scale; (iii) sound levels in decibels — same compression reason; (iv) very large or very small measurements such as the mass of a maize harvest or the diameter of a wire — standard form using powers of ten avoids writing long strings of zeros; (v) pH of soil tested before planting — a logarithmic scale handles concentrations spanning many powers of ten. (Do not award the explanation mark where the candidate merely restates the situation.)

14. Question 14

    3 marks

    Simplify, leaving your answer in index form: $(2^5\times 2^{-3})/2^4$.

    (3 marks)

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