Grade 12 Maths · Functions

Find an exponent using logs

Use logs to find an unknown power, for example solve 2ⁿ = 32 by writing n = log₂ 32 = 5, such as the number of months to save.

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Number of Payments | Gr 12 Financial MathsKevinmathscience

Financial maths Gr12 using logs to find "n" in the formula (mathdou)Dou Kleefstra

Financial Maths - Using Logs to Find n Gr 12Watobe

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Question 7.2 · May/June 2025 · Paper 1 · National (DBE) · 5 marks

7.2 Tino invests

R 500000

in an account earning interest at the rate of

6%

p.a., compounded quarterly. Tino decides to withdraw

R 11250

at the end of every

3

months. Tino will continue making these regular withdrawals until there is no money in the account. How many withdrawals of

R 11250

will Tino be able to make?(5)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: find an exponent using logs

Work each one out first, then open the memo and check every tick.

3. Question 1.1.3 · May/June 2025 · Paper 1 · National (DBE) · 3 marks
1.1 Solve for $x$ :
1.1.3 $2^{x + 4} + 2^{x} = 8704$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Factorise the common exponential term $2^{x} (16 + 1) = 8704$
  1 mark · Take out a common power
2. ✓
  Simplify to a single exponential equation $2^{x} (17) = 8704$
  $2^{x} = 512$
  1 mark · Solve a linear equation
3. ✓
  Solve for x $2^{x} = 2^{9}$
  $x = 9$
  1 mark · Solve an exponential equation
4. Question 5.2.6 · September 2024 · Paper 1 · Provincial · 3 marks
5.2 The graph of $h (x) = a^{x}$ is sketched below. $A (- 1; \frac{1}{2})$ is a point on the graph of h.
5.2.6 If $g (x) = (100) . 3^{x}$ , determine the values of x for which $h (x) = g (x)$ .(3)
Show the memo (3 ticks)Memo
1. ✓
  Simplification $2^{x} = 100 \cdot 3^{x} \Rightarrow (\frac{2}{3})^{x} = 100$
  1 mark · Use the laws of exponents
2. ✓
  Log form $lo g_{\frac{2}{3}} 100 = x$
  1 mark · Find an exponent using logs
3. ✓
  Answer $x = - 11, 36$
  1 mark · Find an exponent using logs
5. Question 7.2.2 · September 2024 · Paper 1 · Provincial · 4 marks
7.2 Micaela buys furniture to the value of R10 000. She borrows the money on 1 February 2023 from a financial institution that charges interest at a rate of 9,5% p.a. compounded monthly. Micaela agrees to pay monthly instalments of R450. The loan agreement allows Micaela to start paying equal monthly instalments from 01 August 2023.
7.2.2 How many months will it take Micaela to pay back the loan?(4)
Show the memo (4 ticks)Memo
1. ✓
  Substitute into Present Value formula $10402, 15 = \frac{450 [ 1 - ( 1 + i ) ^{- n} ]}{i}$
  1 mark · Use the present value formula
2. ✓
  Isolate the bracket $1 - (1, 0079)^{- n} = 0, 183$
  1 mark · Solve a linear equation
3. ✓
  Correct use of logs $0, 8169 = (1, 0079)^{- n}$ $- n = lo g_{1, 0079} 0, 8169$ $n = 25, 63...$
  1 mark · Find an exponent using logs
4. ✓
  Answer $n = 26$ months
  1 mark · Find an exponent using logs
6. Question 1.2.3 · March 2022 · Provincial · 5 marks
1.2 Given: $\sum_{k = 1}^{\infty} 4 (0, 2)^{k - 1}$
1.2.3 Hence calculate the smallest number of terms of the series whose sum will differ by less than $0, 0001$ from the sum to infinity of the series.(5)
Show the memo (5 ticks)Memo
1. ✓
  Set up inequality $S_{\infty} - S_{n} < 0, 0001$
  $5 - \frac{4 ( 1 - ( 0 , 2 ) ^{n} )}{1 - 0 , 2} < 0, 0001$
  1 mark · Calculate a geometric series sum
2. ✓
  Simplify inequality $5 - 5 (1 - (0, 2)^{n}) < 0, 0001$
  $5 - 5 + 5 (0, 2)^{n} < 0, 0001$
  $5 (0, 2)^{n} < 0, 0001$
  1 mark · Solve a linear inequality
3. ✓
  Solve for n using logs $(0, 2)^{n} < 0, 00002$
  $n > lo g_{0, 2} (0, 00002)$ (Note: sign flips if taking log base < 1, or rearrange)
  Using base 5: $5 (5^{- n}) < 1 0^{- 4} ⟹ 5^{1 - n} < 1 0^{- 4}$
  1 mark · Find an exponent using logs
4. ✓
  Calculate n $n > 6, 7$
  1 mark · Find an exponent using logs
5. ✓
  Final answer $n = 7$
  1 mark · Reject solutions that do not fit

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