International Mathematics 0607 Chapter Notes
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The logarithmic function is the inverse of the exponential function. The exponential a^(x) takes a power and gives a number. The logarithm log_a takes the number and gives the power.
The small number a is the base. All logs are base 10 unless the question says otherwise, so log x means log₁₀ x.
log_a(a^(x)) = x
a^(log_a x) = x
Worked example 11 mark
Write down the value of log₄ 64.
4³ = 64, so log₄ 64 = 3 [1]
A power and a log say the same thing. y = a^(x) is equivalent to x = log_a y. The log is the power.
To change form, read the base, the power and the number. The base stays the base. The power becomes the answer of the log.
Two values are worth knowing: log 10 = 1 and log 100 = 2. A log of a power of 10 is the power, so log 1000 = 3.
y = a^(x) iff x = log_a y
log x = p iff x = 10^(p)
Examiner insight
Worked example 12 marks
Do not use a calculator. Solve log x = -2.
Write in index form: x = 10⁻² [1]
x = 1/100 = 0.01 [1]
To solve a^(x) = b, take logs of both sides. The power x comes down in front of log a, so x log a = log b.
Divide to get x = (log b)/(log a). Put log b on top and log a underneath.
On a graphic display calculator, enter log(b) ÷ log(a) with brackets round each log, then read off the value.
a^(x) = b implies x = (log b)/(log a)
Examiner insight
Worked example 13 marks
Solve 3^(x) = 20. Give your answer correct to 3 significant figures.
Take logs of both sides: x log 3 = log 20 [1]
x = (log 20)/(log 3) [1]
Enter log(20) ÷ log(3) on the calculator: x = 2.73 [1]
Growth by a fixed percentage each period gives a power. A multiplier of 1.04 means a 4% rise each year, so after n years the value is the start value times 1.04ⁿ.
To find when a value is reached, set up start × multiplierⁿ = target. The n is a power, not a number to multiply by. Divide by the start value first. Then take logs.
Check what the question asks for. A number of whole years, or the actual year, changes the last line.
value = start × multiplierⁿ
n = (log(target ÷ start))/(log(multiplier))
Examiner insight
Worked example 14 marks
5000 dollars is invested at 4% compound interest each year. Find the smallest whole number of years for the value to reach 8000 dollars.
Multiplier 1.04: 5000 × 1.04ⁿ = 8000 [1]
1.04ⁿ = 1.6 [1]
n = (log 1.6)/(log 1.04) = 11.98 [1]
Whole years: n = 12 [1]
A fixed percentage fall gives a multiplier below 1. A fall of 15% each year is a multiplier of 0.85, because 85% is left.
The method is the same as for growth. The target is below the start value, so log(target ÷ start) and log 0.85 are both negative. A negative divided by a negative gives a positive n.
The value falls past the target part-way through a period. Round up to the next whole period for the first one below the target.
value = start × multiplierⁿ, 0 < multiplier < 1
Examiner insight
Worked example 14 marks
A machine costs 12 000 dollars and loses 15% of its value each year. Find the number of whole years after which its value is first below 5000 dollars.
Multiplier 0.85: 12 000 × 0.85ⁿ = 5000 [1]
0.85ⁿ = 5/12 [1]
n = (log(5/12))/(log 0.85) = (-0.3802)/(-0.07058) = 5.39 [1]
After 5 years the value is still above 5000 dollars, so round up: 6 years [1]
A log equation can also be solved from a graph. Draw y = a^(x) and the line y = b on the calculator. The x value at the intersection is the solution.
Set the window so that the intersection is on screen. Use the calculator's intersection function and read off the x value.
Show the working for this method too: sketch the two curves and write down the intersection.
Examiner insight
Worked example 13 marks
Solve 2^(x) = 50 using the graphs of y = 2^(x) and y = 50. Give your answer correct to 3 significant figures.
Enter y = 2^(x) and y = 50. Set the window 0 ≤ x ≤ 8 and 0 ≤ y ≤ 100. Sketch both curves. [1]
Use the intersection function: (5.64, 50) [1]
x = 5.64. Check: log(50) ÷ log(2) = 5.64 [1]
Recap
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