Change of Base Formula
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Change of Base Formula — Free MYP5 Mathematics (Extended) Practice Questions

1QuestionConverting between basesAssessment Practice
6 marks~9 minCriterion B
The diagram shows the first four figures of a pattern made from small squares arranged in an L-shape.

Figure 1 has 2 squares, Figure 2 has 5 squares, Figure 3 has 10 squares, and Figure 4 has 17 squares.
a
Write down the number of squares in Figure 5. [1]
b
Find a rule for the number of squares SS in Figure nn, giving your answer in the form S=n2+cS = n^2 + c, where cc is an integer. [2]
c
The total number of squares across two consecutive figures, Figure nn and Figure n+1n+1, is 223. Determine whether n=10n = 10 is a valid figure number for this pattern, justifying your answer with full algebraic working. [3]
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Solutions

2QuestionApplying change of base formulaAssessment Practice
6 marks~9 minCriterion D
A student invests 1000 dollars in two savings accounts.

Account A offers an annual interest rate of 5%, compounded annually.
Account B offers an annual interest rate of 4.8%, compounded quarterly.

The balance after tt years in Account A is A(t)=1000(1.05)tA(t) = 1000(1.05)^t and in Account B is B(t)=1000 ⁣(1+0.0484)4tB(t) = 1000\!\left(1 + \dfrac{0.048}{4}\right)^{4t}.

Observed balances (dollars):

tt (years)51015
Account A actual127616292079
Account B actual126816102045
a
Complete the predicted balances from the models at t=5t = 5, t=10t = 10, and t=15t = 15. Give each answer to the nearest dollar. [1]

tt (years): 5, 10, 15
Predicted A (USD):
Predicted B (USD):
b
The number of years for an account to double is given by
t=log⁡2log⁡(1+reff),t = \frac{\log 2}{\log(1 + r_{\text{eff}})},
where reffr_{\text{eff}} is the effective annual interest rate. For Account B,
reff=(1+0.0484)4−1.r_{\text{eff}} = \left(1 + \frac{0.048}{4}\right)^4 - 1.

Calculate reffr_{\text{eff}} for Account B, and hence find the doubling time for each account. Give your answers to 2 decimal places. [3]
c
The percentage error between predicted and actual balance is
percentage error=∣predicted−actual∣actual×100%.\text{percentage error} = \frac{|\text{predicted} - \text{actual}|}{\text{actual}} \times 100\%.

For Account A at t=10t = 10 the percentage error is approximately 0.06%, and for Account B at t=10t = 10 it is approximately 0.12%. A student claims that because both errors are below 1%, both models are equally reliable for predicting when each balance will reach 2000 dollars.

Assess this claim, referring to the doubling times found in part (b) and one limitation of using these models. [2]
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Solutions

3QuestionApplying change of base formulaAssessment Practice
7 marks~11 minCriterion A
Let f(x)=log⁡3xf(x) = \log_3 x and g(x)=log⁡3(x−2)g(x) = \log_3(x - 2).
a
State the equation of the vertical asymptote of g(x)g(x) and write down its domain. [2]
b
Using the change of base formula
log⁡3a=log⁡alog⁡3,\log_3 a = \frac{\log a}{\log 3},
find the exact value of xx such that f(x)=g(x)+2f(x) = g(x) + 2, showing all steps clearly. [3]
c
A student claims that because g(x)=log⁡3(x−2)g(x) = \log_3(x - 2), the graph of gg is the graph of ff translated 2 units to the right. Justify whether this claim is correct, and state the value of xx at which g(x)=0g(x) = 0. [2]
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Solutions

4QuestionApplying change of base formulaAssessment Practice
8 marks~12 minCriterion C
The diagram shows four points plotted on a coordinate grid, where the horizontal axis represents nn and the vertical axis represents log⁡2n(8)\log_{2^n}(8), for n=1,2,3,4n = 1, 2, 3, 4.

The table below gives the values of log⁡2n(8)\log_{2^n}(8) for n=1,2,3,4n = 1, 2, 3, 4.

nn: 1, 2, 3, 4
log⁡2n(8)\log_{2^n}(8): 3, 32, 1, 343,\ \dfrac{3}{2},\ 1,\ \dfrac{3}{4}
a
Using the change of base formula
log⁡a(b)=log⁡2(b)log⁡2(a),\log_a(b) = \frac{\log_2(b)}{\log_2(a)},
show that log⁡2n(8)=3n\log_{2^n}(8) = \dfrac{3}{n} for any positive integer nn. [2]
b
Describe the pattern in the values of log⁡2n(8)\log_{2^n}(8) as nn increases, and find the value of nn for which log⁡2n(8)=37\log_{2^n}(8) = \dfrac{3}{7}. [3]
c
A student claims: "Since log⁡2n(8)=3n\log_{2^n}(8) = \dfrac{3}{n}, the value of log⁡2n(8)\log_{2^n}(8) can never equal zero, no matter how large nn becomes." Justify whether this claim is correct, and state what value log⁡2n(8)\log_{2^n}(8) approaches as nn grows very large. [3]
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