Using Logarithms to Find n in Finance | IEB Grade 12 Maths

Q: When do I need logarithms in finance?

Whenever the unknown is $n$, the number of periods or payments.

When the unknown is the number of periods, logarithms are the only way to solve the equation. This is where the log definition from functions meets finance, and it often comes with a final, smaller payment at the end of a loan.

Solving for n using logarithms

Growth and decay. Isolate the power, then take logs:

$(1 + i)^{n} = k \Rightarrow n = \frac{l o g k}{l o g ( 1 + i )}$

Example. How long does it take for money to double at 8% p.a. compounded quarterly?

$(1, 02)^{n} = 2 \Rightarrow n = \frac{l o g 2}{l o g 1 , 02} \approx 35, 0 quarters \approx 8, 75 years$

Present value annuities. Rearrange the formula first:

$P = \frac{x [ 1 - ( 1 + i ) ^{- n} ]}{i} \Rightarrow (1 + i)^{- n} = 1 - \frac{P i}{x}$

$n = - \frac{l o g ( 1 - \frac{P i}{x} )}{l o g ( 1 + i )}$

Future value annuities. $(1 + i)^{n} = 1 + \frac{F i}{x}$ , so $n = \frac{lo g ( 1 + \frac{F i}{x} )}{lo g ( 1 + i )}$ .

Interpreting $n$ . If $n$ is not a whole number, there are that many full payments plus a smaller final payment. For depreciation questions asking "when will it be worth less than...", round up.

Final payment. After $k$ full payments, find the balance still owed and grow it for one more period. That is the final payment.

Example. A loan of R500 000 at 10,5% p.a. compounded monthly, repaid at R6 000 a month: $n \approx 149, 94$ , so there are 149 payments of R6 000 and a final payment at month 150 of about R5 630,00.

Key terms

Logarithm: The exponent a base must be raised to; used to solve for $n$ .
Final payment: A smaller last payment that clears the remaining balance.
Number of periods: $n$ , the count of compounding periods or payments.

Common mistakes

Taking logs before isolating the power.
Rounding $n$ down when the question asks when a value is reached or exceeded.
Forgetting the negative sign when solving present value equations.
Treating $n \approx 149, 94$ as 150 equal payments instead of 149 plus a final payment.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

How many years will it take for an investment to double at 8% p.a. compounded quarterly?

Worked solution

Hidden - try the question first

Solution

$P (1, 02)^{n} = 2 P$
$n = \frac{lo g 2}{lo g 1 , 02}$
$\approx 35, 0$ quarters
$\approx 8, 75$ years

Question 2

5 marks total

routine

A car worth R250 000 depreciates at 15% p.a. on the reducing-balance method. After how many complete years will it first be worth less than R100 000?

Worked solution

Hidden - try the question first

Solution

$P (1, 02)^{n} = 2 P$
$(0, 85)^{n} = 0, 4$
$n = \frac{lo g 0 , 4}{lo g 0 , 85}$
$\approx 5, 64$
After 6 complete years

Question 3

7 marks total

complex

A loan of R500 000 is repaid with payments of R6 000 at the end of every month. Interest is 10,5% p.a. compounded monthly.

(a) How many payments of R6 000 will be made? (4)

(b) Calculate the final payment. (3)

Worked solution

Hidden - try the question first

Solution

(a) $500000 = \frac{6 000 [ 1 - ( 1 , 00875 ) ^{- n} ]}{0 , 00875}$
$(1, 00875)^{- n} = 1 - \frac{500 000 \times 0 , 00875}{6 000} = 0, 270833 \dots$
$n = - \frac{lo g 0 , 270833}{lo g 1 , 00875} \approx 149, 94$
149 payments of R6 000
(b) Balance after 149 payments: $500000 (1, 00875)^{149} - \frac{6 000 [ ( 1 , 00875 ) ^{149} - 1 ]}{0 , 00875} \approx 5581, 17$
Final payment one month later: $5581, 17 \times 1, 00875$
$\approx$ R5 630,00

Question 4

9 marks total

problem-solving

Mrs Dlamini retires with R3 000 000. She invests it at 7,5% p.a. compounded monthly and withdraws R30 000 at the end of every month.

(a) How many full withdrawals can she make? How many years is this? (5)

(b) What equal monthly withdrawal would make the money last exactly 25 years? (4)

Worked solution

Hidden - try the question first

Solution

(a) $500000 = \frac{6 000 [ 1 - ( 1 , 00875 ) ^{- n} ]}{0 , 00875}$
$(1, 00625)^{- n} = 1 - \frac{3 000 000 \times 0 , 00625}{30 000} = 0, 375$
$n = - \frac{lo g 0 , 375}{lo g 1 , 00625}$
$\approx 157, 42$ , so 157 full withdrawals
About 13 years
(b) $n = 300$
$x = \frac{3 000 000 \times 0 , 00625}{1 - ( 1 , 00625 ) ^{- 300}}$
$\approx$ R22 169,74

FAQs

When do I need logarithms in finance?

Whenever the unknown is $n$ , the number of periods or payments.

Should I round n up or down?

It depends on the question. For full payments, round down and add a final payment. For 'when will it reach' questions, round up.

How do I calculate the final payment on a loan?

Find the balance after the last full payment, then add one period of interest. That amount is the final payment.

Using logarithms to find the number of periods

Solving for n using logarithms

Key terms

Common mistakes

Practice questions

FAQs

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