Lesson 5: Exponential and Logarithmic Functions

An exponential function f(x) = a·b^x (a ≠ 0, b > 0, b ≠ 1) grows or decays by a constant percentage per unit step — the model for populations, investments, radioactivity and cooling. Its inverse, the logarithm, answers the reverse question: "how long until?" This lesson defines both, builds fluency with the log laws, and solves equations by moving between the two forms. Every law is stated with its domain, because most logarithm errors are really domain errors in disguise.

Exponential functions and their graphs

For b > 1, f(x) = b^x is increasing (growth); for 0 < b < 1 it is decreasing (decay). All such graphs share: domain all reals, range y > 0, y-intercept (0, 1) (or (0, a) with a coefficient), and the horizontal asymptote y = 0. The base e ≈ 2.71828 — the limit of (1 + 1/n)^n — is the natural choice for continuous growth and calculus. Transformations behave as in Lesson 2: f(x) = 3·2^(x − 1) − 4 has y-intercept 3·2^(−1) − 4 = −2.5, asymptote y = −4, and range y > −4.

The two rows are mirror images: (1/2)^x = 2^(−x), decay read backwards is growth.

Logarithms: the inverse viewpoint

Definition: log_b(y) = x means b^x = y, valid for y > 0. So log_2 32 = 5 because 2^5 = 32; log_5 (1/25) = −2 because 5^(−2) = 1/25; ln(e^3) = 3. The two workhorse bases are log = log_10 (common) and ln = log_e (natural). A logarithm is an exponent: every log statement is an exponential statement rewritten, and converting between the forms is the master skill of this lesson.

Worked Example 1 — evaluating. (a) log_3 81 = 4 since 3^4 = 81. (b) log_16 4 = 1/2 since 16^(1/2) = 4 — fractional exponents give fractional logs. (c) log_7 1 = 0 for every base, since b^0 = 1. (d) log_b b = 1.

The laws of logarithms

For positive M, N and any real r:

• Product: log_b(MN) = log_b M + log_b N
• Quotient: log_b(M/N) = log_b M − log_b N
• Power: log_b(M^r) = r·log_b M
• Change of base: log_b M = (log M)/(log b) = (ln M)/(ln b) — this is how any base is computed on a calculator.

Worked Example 2 — expanding and condensing. (a) Expand: log_2(8x^3/y) = log_2 8 + log_2(x^3) − log_2 y = 3 + 3 log_2 x − log_2 y. (b) Condense: 2 ln(x + 1) − ln x = ln((x + 1)^2/x). (c) Change of base: log_3 50 = ln 50/ln 3 ≈ 3.912/1.099 ≈ 3.56. Check: 3^3.56 ≈ 50 ✓.

Solving exponential and logarithmic equations

Two strategies dominate. For exponentials: rewrite both sides with a common base if possible; otherwise take logs of both sides and use the power law to bring the exponent down. For logarithms: condense to a single log, convert to exponential form, solve — and then check every candidate against the domain (arguments must be positive).

Worked Example 3 — exponential equations. (a) 3^(2x − 1) = 81 = 3^4 gives 2x − 1 = 4, so x = 5/2. (b) 5·2^x = 320: divide by 5, 2^x = 64 = 2^6, so x = 6. (c) 4^x = 30 has no tidy base: x = log 30/log 4 ≈ 1.477/0.602 ≈ 2.45.

Worked Example 4 — a logarithmic equation with a trap. Solve log_2 x + log_2(x − 2) = 3. Condense: log_2(x(x − 2)) = 3, so x(x − 2) = 2^3 = 8, giving x^2 − 2x − 8 = 0 = (x − 4)(x + 2), candidates x = 4 and x = −2. But log_2(−2) is undefined, so x = −2 is extraneous — the product law was applied where its hypotheses failed. Check x = 4: log_2 4 + log_2 2 = 2 + 1 = 3 ✓. The only solution is x = 4. Lessons 10 and 11 return to extraneous solutions; here the cause is the log domain, not squaring.

Modelling: decibels, pH and Richter — why the world uses log scales

Quantities spanning many orders of magnitude are measured on log scales. Sound level is L = 10 log_10(I/I_0) decibels, where I_0 = 10^(−12) W/m^2 is the threshold of hearing. A sound 1,000 times the reference intensity has L = 10 log_10(1000) = 30 dB. Ordinary conversation (≈ 60 dB) versus a whisper (≈ 30 dB): the 30 dB gap means a factor of 10^(30/10) = 1,000 in intensity — the log scale compresses a thousandfold range into a friendly number. Acidity: pH = −log_10[H^+]; a solution of pH 5.5 has [H^+] = 10^(−5.5) ≈ 3.16 × 10^(−6) mol/L, and each pH unit is a tenfold change in concentration — lemon juice at pH 2 is about 1,000 times as acidic as a pH 5 solution. Earthquakes: the Richter magnitude is a base-10 log of amplitude, so magnitude 7 versus magnitude 5 is 10^2 = 100 times the ground motion and roughly 10^(1.5·2) = 1,000 times the energy. Logarithms are not a classroom ornament; they are how science reports anything that multiplies.

Common mistakes

• Inventing a sum law. log(a + b) has no expansion. Counterexample: log_10(10 + 90) = log_10 100 = 2, but log_10 10 + log_10 90 ≈ 1 + 1.954 = 2.954. The product law is about multiplication inside the log, not addition.
• The reverse slip: writing log a + log b = log(a + b). It equals log(ab).
• Power-law direction. log(x^3) = 3 log x, but (log x)^3 is something else entirely; the exponent must be inside the argument to come down.
• Skipping the domain check in log equations. Worked Example 4 produced an algebraic candidate that the domain rejects. Always test candidates in the original equation.
• Treating log as a factor to cancel. In log_2(x) = 5, you cannot "divide by log"; convert to exponential form: x = 2^5 = 32.

Exam Experience

1. log_4 64 is equal to

• (A) 3
• (B) 8
• (C) 16
• (D) 3/2

Answer: (A). 4^3 = 64. (D) is the classic slip of answering log_64 4 or halving the wrong way: check by converting — 4^(3/2) = 8 ≠ 64.

2. The solution of 3^(x − 2) = 27 is

• (A) x = 3
• (B) x = 5
• (C) x = 9
• (D) x = 1

Answer: (B). 27 = 3^3, so x − 2 = 3 and x = 5. Choice (A) equates the exponent with 3 and forgets the −2 shift.

3. log_5(x^2) = 4 has solutions

• (A) x = 25 only
• (B) x = 625 only
• (C) x = 25 and x = −25
• (D) x = 5 and x = −5

Answer: (C). x^2 = 5^4 = 625, so x = ±25. Both are valid because the argument x^2 is positive for both — unlike log_5(x)^... the trap is dropping the negative root as if the argument were x itself. (A) makes exactly that error.

Constructed response. A jackhammer produces sound intensity I = 10^(−2) W/m^2. The decibel level is L = 10 log_10(I/10^(−12)). (a) Find the jackhammer's decibel level. (b) Workplace rules require hearing protection above 85 dB. Find the greatest intensity, in W/m^2, permitted without protection, to three significant figures. (c) Two identical jackhammers run together, doubling the intensity. By how many decibels does the level rise? Give an exact answer.

Scoring guide. (a) 1 point: L = 10 log_10(10^(−2)/10^(−12)) = 10 log_10(10^10) = 100 dB. (b) 2 points: 1 for setting 85 = 10 log_10(I/10^(−12)) and isolating log_10(I/10^(−12)) = 8.5; 1 for I = 10^(−12) · 10^8.5 = 10^(−3.5) ≈ 3.16 × 10^(−4) W/m^2. (c) 2 points: 1 for using the product/quotient law correctly — the rise is 10 log_10(2I) − 10 log_10(I) = 10 log_10 2; 1 for the exact answer 10 log_10 2 ≈ 3.01 dB, with the observation that doubling intensity always adds the same ≈ 3 dB whatever the starting level.

Going deeper: natural logarithms and log graphs

Equations mixing ln with polynomials use the same condense-and-convert method. Solve ln x + ln(x + 3) = ln 10, domain x > 0. Condense: ln(x(x + 3)) = ln 10, so x^2 + 3x = 10, giving x^2 + 3x − 10 = 0 = (x + 5)(x − 2). Candidates 2 and −5; the domain kills −5 (ln(−5) is undefined, and so is ln(−5 + 3) = ln(−2)). Check x = 2: ln 2 + ln 5 = ln 10 ✓ by the product law itself. Notice the aesthetic: the equation was built from the law it tests.

Graphs of logarithms are reflections of exponential graphs (Lesson 2's inverse symmetry). Take y = ln(x − 2) + 1: it is y = ln x shifted right 2 and up 1. Consequences read off without computation: domain x > 2; vertical asymptote x = 2; range all reals; the point (3, 1) (where the argument is 1 and ln 1 = 0); and the x-intercept where ln(x − 2) = −1, i.e. x − 2 = e^(−1) = 1/e, x ≈ 2.37. Every logarithm graph passes through exactly one x-intercept and climbs without bound but ever more slowly — doubling the argument adds a constant ln 2 ≈ 0.693 to the value, however large x already is. That flattening is why log scales (Lesson 5's decibels and pH) can display hurricanes and breezes on the same axis, and why Lesson 18 reaches for logarithms when data spans orders of magnitude.

One more pattern: logs of quotients collapse

Given log_2 96 ≈ 6.585 and log_2 3 ≈ 1.585, the quotient law gives log_2(96/3) = log_2 32 = 5 exactly — the approximations subtract to 5.000, and the exact answer was available all along because 96/3 = 32 is a power of 2. Train the reflex: before reaching for decimals, look for the power. The same reflex evaluates log_5 125 − log_5 5 = log_5 25 = 2, and in reverse it powers the inverse identities 10^(log x) = x and e^(ln x) = x (for x > 0): a function and its inverse cancel, leaving the input untouched — Lesson 2's f⁻¹(f(x)) = x in its most-used special case.