Cambridge International AS & A Level Mathematics 9709 (2026-2027) · Paper 2: Pure Mathematics 2 · official syllabus
LOGS AND EXPONENTIALS LAB
Learn it by playing it
Four short missions on the laws of logarithms, the curve y = e^(kx), solving equations with the unknown in an index, and straightening power laws on ln-ln axes. When a mission glows green, read why it worked, then take the next one.
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Mission 1 of 4
Pull the powers out front
The target bar shows the value of ln(5^3 x 7^2). Your bar shows p ln 5 + q ln 7. Use the law ln(u^n) = n ln u to rebuild the target exactly - slide p and q until the bars coincide.
Before you touch a slider: ln(a^3 b^2) equals which expression?
Your bar: 1 ln 5 + 1 ln 7 = 3.5553. Target: ln(5^3 x 7^2) = 8.7201. Bring the 3 down in front of ln 5. Bring the 2 down in front of ln 7.
Study notes
Played the lab above? These notes are the reference version of what you just felt - use them to check exact formulas and the exam wording.
Logarithms are indices read backwards 2.2
The statements a^x = b and x = log_a(b) say the same thing: the logarithm is the power that produces b from base a. So 2^5 = 32 and log_2(32) = 5 are one fact, not two.
The natural logarithm ln x is the logarithm to base e, where e = 2.71828... is the constant whose exponential function is its own derivative. ln(e^x) = x and e^(ln x) = x: the two functions undo each other.
A logarithm only takes positive inputs, because a positive base raised to any real power stays positive. ln(-2) has no real value, and a solution that asks for ln of a nonpositive number is invalid.
Handy values: ln 1 = 0 (any base to the power 0 is 1) and ln e = 1. These two points anchor every sketch.
The laws of logarithms 2.2
Product to sum: ln(uv) = ln u + ln v. Quotient to difference: ln(u/v) = ln u - ln v. Power out front: ln(u^n) = n ln u. Each law is the matching index law read through the definition of a logarithm.
The laws combine: ln(a^3 b^2) = 3 ln a + 2 ln b, and ln(50/7) = ln 50 - ln 7. Use them to compress several logs into one, or to split one log apart - whichever direction the question needs.
There is no law for ln(u + v): the log of a sum does not split. ln(a + b) is not ln a + ln b, and ln(u^n) is not (ln u)^n.
Change of base is explicitly excluded from 9709 Paper 2. Every equation can be handled with natural logarithms alone, so stay in ln throughout.
e^x and ln x: properties and graphs 2.2
y = e^x passes through (0, 1), is always positive, and grows ever faster as x increases; as x decreases it decays towards 0. The x-axis is a horizontal asymptote on the left.
y = ln x is the mirror of y = e^x in the line y = x: it passes through (1, 0), is defined only for x > 0, increases throughout, and drops without bound as x approaches 0 from the right.
For y = e^(kx) with k > 0 the graph is growth, steeper for larger k; with k < 0 it is decay, falling towards 0. Every one of these curves passes through (0, 1) because e^0 = 1.
Because e^x and ln x are inverse functions, applying one after the other returns the input: e^(ln 7) = 7 and ln(e^7) = 7. This inverse pair is what unlocks equations with the unknown in an index.
Solving equations with the unknown in an index 2.2
Take ln of both sides, bring the power down, then isolate x. For 2^(x+1) = 5: (x + 1) ln 2 = ln 5, so x = ln 5 / ln 2 - 1, which is about 1.322.
When both sides have the unknown in an index, the same move works: 3^x = 4^(2x-1) gives x ln 3 = (2x - 1) ln 4, so x(2 ln 4 - ln 3) = ln 4 and x = ln 4 / (2 ln 4 - ln 3).
A quotient of logs is not the log of a quotient: ln 10 / ln 4 is a single number (about 1.661), while ln(10/4) and ln 10 - ln 4 are different numbers (about 0.916). Mixing these up is the most common slip.
Give the exact form first (x = ln 10 / ln 4), then the decimal if the question asks. Check by substitution: 4^1.661 is close to 10, and small rounding in x moves 4^x a lot.
Inequalities with the unknown in an index 2.2
ln is an increasing function, so taking ln of both sides of an inequality between positive quantities keeps the direction: 2^x > 7 gives x ln 2 > ln 7, so x > ln 7 / ln 2.
For a two-sided bound, work on both ends: 3 <= 2^(3x+1) <= 5 gives ln 3 <= (3x + 1) ln 2 <= ln 5, so (ln 3 / ln 2 - 1)/3 <= x <= (ln 5 / ln 2 - 1)/3.
Watch for bases between 0 and 1 in disguised form: 0.5^x is decreasing in x, so after taking ln the division by ln 0.5 (a negative number) reverses the inequality. Taking ln of both sides directly avoids this trap because ln keeps the direction and the sign of ln b handles the rest.
State the solution as an interval or inequality in x, and check a test value inside and outside it against the original inequality.
Linear form: straightening data to find constants 2.2
A power law y = kx^n becomes linear after taking ln: ln y = n ln x + ln k. Plotting ln y against ln x gives a straight line whose gradient is the power n and whose intercept is ln k.
An exponential law y = k(a^x) also becomes linear: ln y = (ln a) x + ln k. Plot ln y against x itself; the gradient is ln a and the intercept is ln k.
To find the constants from two points, substitute them into the linear equation. Points (1, 2) and (2, 16) on y = kx^n give ln y values ln 2 and ln 16 at ln x values 0 and ln 2: the gradient is (ln 16 - ln 2)/(ln 2 - 0) = 3 and the intercept is ln 2, so n = 3 and k = 2.
Convert back at the end: an intercept of ln k means k = e^(intercept), and a gradient of ln a means a = e^(gradient). Report k, n or a as ordinary numbers unless the exact log form is requested.
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