4PM1/01 January 2023 Question 7: Exponentials, Logarithms, e and ln

Question 1

Fill in the missing y-values in the table for y = 0.5^(x/3 + 1) + 2, rounding to 2 decimal places where needed.

Accepted Answer

x = −3: 0.5⁰ + 2 = 3 x = −2: 0.5^(1/3) + 2 = 2.79 x = −1: 0.5^(2/3) + 2 = 2.63 Answer: x = −3: y = 3; x = −2: y = 2.79; x = −1: y = 2.63

Question 2

Draw the graph of y = 0.5^(x/3 + 1) + 2 on the grid for x from −6 to 0.

Accepted Answer

Plot the seven points Join them with a smooth curve Answer: Points from the completed table plotted and joined with a smooth decreasing curve from (−6, 4) to (0, 2.5)

Question 3

Draw a suitable straight line on the grid and use it to estimate, to one decimal place, the root of log₂(2x + 2)³ + x + 3 = 0 for −6 ≤ x ≤ 0.

Accepted Answer

log₂ (2x + 2)³ = 3 log₂ 2(x + 1) = 3 + 3 log₂ (x + 1) So 3 log₂ (x + 1) + x + 6 = 0, log₂ (x + 1) = −(x/3 + 2) x + 1 = 2^(−(x/3 + 2)) = 0.5 × 0.5^(x/3 + 1) 0.5^(x/3 + 1) = 2x + 2, so 0.5^(x/3 + 1) + 2 = 2x + 4 Draw y = 2x + 4; it meets the curve at x ≈ −0.7 Answer: Draw y = 2x + 4; x ≈ −0.7

4PM1/01 Jan 2023, Question 7: Exponentials, Logarithms, e and ln

The question, in short

Part (a) · 2 marks

Part (b) · 2 marks

Part (c) · 6 marks

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