On a semi-log plot the x-axis is ordinary, but the y-axis is logarithmic: the spacing of tick marks depends on log y, so each step up multiplies the value instead of adding to it. This lets you display data that cover a huge range of values. The key fact is that an exponential function becomes a straight line on a semi-log plot. To see why, take a log of both sides of y = a·bx: log y = x·log b + log a. This is a line with slope log b and intercept log a. Similarly, for y = a·ekx, ln y = kx + ln a. This process is called linearizing. If data points fall on a line on a semi-log plot, an exponential model fits well, and the line’s slope and intercept give you b and a.
A logarithm answers one question: what exponent do I put on the base to get this number? The expression logb c, read “log base b of c,” is the exponent a that makes ba = c. The two statements are the same fact in two forms: logb c = a means ba = c, where b > 0, b ≠ 1, and c > 0. When the base is omitted, as in log x, it is understood to be 10 (a common logarithm), so log 1,000 = 3 because 103 = 1,000. Some values can be found by thinking in powers, and others need a calculator. If your calculator only has the LOG (base 10) and LN (base e) keys, use the change of base formula logb x = log x / log b (or ln x / ln b, which gives the same result). A logarithm of a number between 0 and 1 is negative, logb 1 = 0 for every base, and logb b = 1.
Worked Example: Evaluate (a) log3 81, (b) log7 √7, (c) log2(1/16), (d) log 1,000, and (e) log6 20 to the nearest hundredth.
(a) 34 = 81, so the answer is 4.
(b) √7 = 71/2, so the answer is 1/2.
(c) 1/16 = 2−4, so the answer is −4.
(d) 1,000 = 103, so the answer is 3.
(e) 20 is not a nice power of 6, so use the change of base formula: ln 20 / ln 6 ≈ 2.9957 / 1.7918 ≈ 1.67.
4, 1/2, −4, 3, and about 1.67
1. What is log4 64?
Explanation: Ask what power of 4 gives 64: 43 = 64, so log4 64 = 3.
2. What is log2(1/8)?
Explanation: 1/8 = 2−3, so the exponent is −3. A number between 0 and 1 always has a negative logarithm when the base is greater than 1.
3. Which equation is the logarithmic form of 53 = 125?
Explanation: In logb c = a, the base b stays the base, the answer a is the exponent, and c is the result. Here b = 5, a = 3 and c = 125.
4. What is log5 40, to the nearest hundredth?
Explanation: Change of base: log 40 / log 5 = 1.6021 / 0.6990 ≈ 2.29. The value 3.69 is ln 40, and 1.60 is log 40, which are not log5 40.
The logarithmic function y = logb x is defined as the inverse of the exponential function y = bx, so y = logb x means by = x. Because they are inverses, each one undoes the other. This gives the two identities you will use constantly: logb(bx) = x for every real x, and blogb x = x for every x > 0. In words, the log “cancels” a power of the same base, and a power “cancels” a log of the same base. The identities also hold for common logs (10log x = x, log 10x = x) and natural logs (eln x = x, ln ex = x). Watch the base: the identity only works when the base of the log and the base of the power match.
Worked Example: Simplify each expression.
(a) log7 7x+2 = x + 2, since the log cancels the power of 7.
(b) 5log5 12 = 12, since the power cancels the log of the same base.
(c) 102 log 3: bring the 2 up as a power, 2 log 3 = log 9, so 10log 9 = 9.
(d) ln e−3 = −3.
x + 2, 12, 9, −3
1. Simplify log2(25x).
Explanation: log2 cancels the base-2 power, leaving the exponent 5x.
2. What is 6log6 17?
Explanation: A power of 6 and a log base 6 undo each other, so the result is the number inside the log, 17.
3. What is e2 ln 3?
Explanation: First rewrite 2 ln 3 = ln 9. Then eln 9 = 9.
4. If logb 49 = 2 and b > 0, what is b?
Explanation: Rewrite in exponential form: b2 = 49. Since b > 0, b = 7.
Because y = logb x is the inverse of y = bx, its graph is the reflection of the exponential graph over the line y = x, and every ordered pair (s, t) of the exponential becomes (t, s) for the logarithm. So to graph y = logb x, make a table for bx and swap the columns. The inputs and outputs swap roles, so the domain and range swap too. Summary of y = logb x:
Worked Example: Use y = 2x to graph y = log2 x, and state the domain and range of each.
Table for 2x: (−3, 1/8), (−2, 1/4), (−1, 1/2), (0, 1), (1, 2), (2, 4), (3, 8).
Swap the coordinates for log2 x: (1/8, −3), (1/4, −2), (1/2, −1), (1, 0), (2, 1), (4, 2), (8, 3).
y = 2x: domain (−∞, ∞), range (0, ∞). y = log2 x: domain (0, ∞), range (−∞, ∞).
The graphs are mirror images over y = x
1. The point (−1, 1/5) is on y = 5x. Which point is on y = log5 x?
Explanation: Inverse functions swap coordinates, so (−1, 1/5) becomes (1/5, −1). Check: 5−1 = 1/5, so log5(1/5) = −1.
2. What are the domain and range of y = log3 x?
Explanation: You can only take the log of a positive number, so x > 0. The outputs are exponents, which can be any real number.
3. What are the x-intercept and vertical asymptote of y = log7 x?
Explanation: log7 1 = 0, so (1, 0) is on the graph. The graph approaches the y-axis, x = 0, without touching it.
4. Which describes y = log1/3 x?
Explanation: For a base between 0 and 1 the exponential decays, and so does its inverse. Every logarithm passes through (1, 0).
Logarithmic functions have the vertical asymptote x = 0 and an unbounded end behavior, so every limit at the ends is ∞ or −∞. For y = logb x with b > 1: limx→∞ = ∞, limx→0+ = −∞, and limx→1 = 0. With 0 < b < 1 the graph is flipped, so the two infinite limits swap signs. A multiplier a in y = a·logb x also flips the signs when a is negative. Only the right-hand limit x → 0+ exists, because the function is not defined for x ≤ 0. Although the graph keeps rising, it does so very slowly.
Worked Example: Let f(x) = log4 x and g(x) = log1/5 x. Evaluate limx→∞, limx→0+, and limx→1 of each.
f has a base greater than 1: limx→∞ f = ∞, limx→0+ f = −∞, limx→1 f = log4 1 = 0.
g has a base between 0 and 1: limx→∞ g = −∞, limx→0+ g = ∞, limx→1 g = 0.
f: ∞, −∞, 0. g: −∞, ∞, 0.
1. What is limx→0+ log2 x?
Explanation: As x gets closer to 0 from the right, the exponent needed to make 2 to that power small grows more and more negative.
2. What is limx→∞ log1/2 x?
Explanation: The base is between 0 and 1, so the function decreases as x increases and has no lower bound.
3. What is limx→∞ (−3 log5 x)?
Explanation: log5 x → ∞, and multiplying by −3 makes the outputs grow without bound in the negative direction.
4. What is limx→0+ (−3 log5 x)?
Explanation: log5 x → −∞ as x → 0+, and multiplying by −3 flips the sign, giving ∞.
Because a logarithm is an exponent, the logarithm laws mirror the exponent laws. For positive x and y and any valid base:
Worked Example: If log 2 = p and log 5 = q, express (a) log 40 and (b) log √(5/2) in terms of p and q.
(a) 40 = 23·5, so log 40 = 3 log 2 + log 5 = 3p + q.
(b) log √(5/2) = (1/2)(log 5 − log 2) = (q − p)/2.
Condense example: 2 log x − (1/2) log y = log x2 − log y1/2 = log(x2/√y).
(a) 3p + q; (b) (q − p)/2
1. If log a = 3 and log b = 1, what is log(a2/√b)?
Explanation: log(a2/√b) = 2 log a − (1/2) log b = 2(3) − (1/2)(1) = 6 − 0.5 = 5.5.
2. Which single logarithm equals 2 log x − (1/2) log y?
Explanation: The power law moves coefficients up as exponents: 2 log x = log x2 and (1/2) log y = log √y. The difference is then the log of a quotient.
3. If log N = (1/2)(log a + log b) − log c, what is N?
Explanation: (1/2)(log a + log b) = (1/2) log(ab) = log √(ab). Subtracting log c gives log(√(ab)/c), so N = √(ab)/c. Choice C wrongly turns the sum of logs into a sum of numbers.
4. If log 50 = b, which expression equals b − 1?
Explanation: Since 1 = log 10, b − 1 = log 50 − log 10 = log(50/10) = log 5. Subtracting 1 from a log does not mean subtracting 1 from the number.
The laws explain how the graph of a log function is transformed. Every horizontal dilation becomes a vertical translation, because logb(kx) = logb k + logb x. Raising the input to a power becomes a vertical dilation, since logb(xn) = n logb x. And by change of base, all logarithmic functions are vertical dilations of each other: logb x = (1/loga b)·loga x. To describe a transformation, first rewrite the function as a·logb x + k in the same parent base, then read off the dilation and shift. The natural logarithm ln x is loge x, with e ≈ 2.718. It has the same laws, plus ln e = 1, ln ex = x, and eln x = x. Use the LN key for values such as ln 5 ≈ 1.6094.
Worked Example: Describe each function as a transformation of its parent.
(a) f(x) = log2(8x) = log2 8 + log2 x = 3 + log2 x: the graph of y = log2 x moved up 3.
(b) g(x) = log3 x2 = 2 log3 x: a vertical stretch by a factor of 2 of y = log3 x.
(c) h(x) = log9 x = log3 x / log3 9 = (1/2) log3 x: a vertical shrink by a factor of 1/2 of y = log3 x.
(d) Natural log: ln 5 ≈ 1.6094, 2 ln 3 = ln 9 ≈ 2.1972, ln e = 1, ln e−4 = −4.
1. The graph of y = log2(8x) can be obtained from y = log2 x by:
Explanation: log2(8x) = log2 8 + log2 x = 3 + log2 x, which is a vertical shift of 3 up.
2. Compared with y = log5 x, the graph of y = log5(x2) is:
Explanation: log5 x2 = 2 log5 x, which multiplies every output by 2.
3. Compared with y = log2 x, the graph of y = log4 x is:
Explanation: log4 x = log2 x / log2 4 = (1/2) log2 x.
4. What is e3 ln 2?
Explanation: 3 ln 2 = ln 23 = ln 8, so e3 ln 2 = eln 8 = 8.
5. If f(x) = log(x3) and g(x) = x/100, which expression equals f(g(x))?
Explanation: f(g(x)) = log((x/100)3) = 3 log(x/100) = 3(log x − log 100) = 3(log x − 2).
Logarithmic equations. If every term is a log with the same base, use the laws to write each side as a single log, then use the one-to-one property: if logb A = logb B, then A = B. If the equation has a constant, collect all the logs on one side, condense into one log, rewrite in exponential form, and solve. Always check every answer in the original equation and reject any that puts a zero or negative number inside a log; those are extraneous solutions. Exponential equations. If both sides cannot be written with the same base, take a log of both sides and use the power law to bring the variable down: from bkx = c get kx·log b = log c, so x = log c / (k log b). For models such as f(t) = a·ekt, take ln of both sides.
Worked Example: (a) Solve log3 x + log3(x − 2) = 1. (b) Solve 32x = 20.
(a) Condense: log3[x(x − 2)] = 1, so x(x − 2) = 31. Then x2 − 2x − 3 = 0, so (x − 3)(x + 1) = 0 and x = 3 or x = −1. For x = −1, log3(−1) is undefined, so reject it. Check x = 3: log3 3 + log3 1 = 1 + 0 = 1 ✓.
(b) Take ln of both sides: 2x ln 3 = ln 20, so x = ln 20 / (2 ln 3) ≈ 2.9957 / 2.1972 ≈ 1.363.
(a) x = 3 only; (b) x ≈ 1.363
1. If log2(x2 + 3) = 3, what is x?
Explanation: Exponential form: x2 + 3 = 23 = 8, so x2 = 5 and x = ±√5. Both are valid because x2 + 3 is positive for every real x, so neither is extraneous.
2. What is the solution of log2 x + log2(x − 2) = 3?
Explanation: Condense: x(x − 2) = 8, so x2 − 2x − 8 = 0, giving (x − 4)(x + 2) = 0. x = −2 puts a negative number inside log2 x, so it is extraneous.
3. Solve 5x = 12. Round to the nearest hundredth.
Explanation: x = ln 12 / ln 5 = 2.4849 / 1.6094 ≈ 1.54. Check: 51.54 ≈ 12.
4. A 50 mg dose falls continuously, and after 3 hours 40 mg remain. Using f(t) = 50ekt, about how many hours until half the dose (25 mg) remains?
Explanation: From 40 = 50e3k, e3k = 0.8, so k = ln 0.8 / 3 ≈ −0.0744. Then 25 = 50ekt gives ekt = 0.5, so t = ln 0.5 / k ≈ (−0.6931)/(−0.0744) ≈ 9.3.
The methods for equations carry over to inequalities, with two extra cautions. First, the domain: inputs of a log must be positive, so always intersect your answer with the domain. Second, the direction of the base: if the base is greater than 1 the function is increasing, so the inequality direction is kept when you drop the logs or rewrite in exponential form. If the base is between 0 and 1 the function is decreasing, so the inequality reverses. A reliable plan is to solve the related equation first, then use the domain and a test value in each region (or the graph) to decide which side to keep.
Worked Example: Solve (a) 2x > 50, (b) log x < 3, and (c) log2(3x − 1) > log2(x + 5).
(a) Take logs: x log 2 > log 50, so x > log 50 / log 2 ≈ 1.6990 / 0.3010 ≈ 5.644.
(b) Solve log x = 3 to get x = 103 = 1000. Since the input must be positive, the solution is 0 < x < 1000.
(c) The base 2 is greater than 1, so 3x − 1 > x + 5, which gives x > 3. The domain needs 3x − 1 > 0 and x + 5 > 0, which holds for x > 3, so nothing is removed.
(a) x > 5.644; (b) 0 < x < 1000; (c) x > 3
1. Solve 3x ≥ 20. Round to the nearest hundredth.
Explanation: x ≥ ln 20 / ln 3 = 2.9957 / 1.0986 ≈ 2.73. The base is greater than 1, so the direction does not change.
2. Solve log3 x ≤ 2.
Explanation: log3 x = 2 gives x = 9. Since the input must be positive, the solution is 0 < x ≤ 9.
3. Solve log2(2x + 1) > log2(x + 4).
Explanation: With base 2 the direction is kept: 2x + 1 > x + 4, so x > 3. The domain conditions x > −1/2 and x > −4 are already satisfied.
4. Solve log1/2 x > 1.
Explanation: The base 1/2 is less than 1, so the function is decreasing and the inequality reverses when you rewrite: x < (1/2)1. With x > 0, the solution is 0 < x < 1/2.
A transformed logarithm has the form f(x) = a·logb(x − h) + k. The horizontal shift moves the vertical asymptote to x = h and the domain to x > h, and k shifts the graph vertically. The easiest way to graph it is to track the parent’s anchor point (1, 0): it moves to (1 + h, k). Reflection and dilations come from a. To find the domain of any log expression, set the input of the log greater than 0. To find the inverse of a transformed function, swap x and y, isolate the log (or the exponential), rewrite in the other form, and solve for y. The inverse of a log is an exponential and vice versa, and the asymptote of one becomes the asymptote of the other, rotated (vertical ↔ horizontal).
Worked Example: For g(x) = log2(x − 1) + 4, describe the graph, state the domain, and find g−1(x).
The graph of y = log2 x moves right 1 and up 4, so the point (1, 0) moves to (2, 4) and the asymptote is x = 1. The domain needs x − 1 > 0, so x > 1.
Inverse: swap to x = log2(y − 1) + 4. Then x − 4 = log2(y − 1), so y − 1 = 2x−4, and y = 2x−4 + 1.
Domain x > 1; g−1(x) = 2x−4 + 1
1. What is the domain of f(x) = log(x − 3)?
Explanation: The input of the log must be positive: x − 3 > 0, so x > 3.
2. Which function is the inverse of y = log(x − 2)?
Explanation: Swap: x = log(y − 2), so y − 2 = 10x and y = 10x + 2.
3. What is the vertical asymptote of y = log4(x + 5) − 2?
Explanation: The input is zero when x + 5 = 0, so the asymptote is x = −5. The −2 only shifts the graph vertically.
4. What is the inverse of f(x) = 3x − 4?
Explanation: Swap: x = 3y − 4, so 3y = x + 4 and y = log3(x + 4). Check: f(2) = 5 and f−1(5) = log3 9 = 2.
Logarithmic functions model quantities that grow quickly at first and then slow down but never stop increasing, such as a learning curve, or any quantity where equal ratios of the input give equal differences in the output (the Richter and decibel scales, for example, are logarithmic because the intensity is an exponential function of the reading). To build a model y = a + b ln x (or a·logb x + k) from points, use a point with x = 1, where ln 1 = 0, to find the constant a, and then a second point to find b. With data and a calculator, use logarithmic regression (LnReg), which returns a and b for y = a + b ln x. Using a model inside the range of the data is interpolation and is generally reliable. Using it outside the range is extrapolation and is much less certain.
Worked Example: A skill score rises quickly and then levels, and passes through (1, 4) and (5, 9). Build the model y = a + b ln x and predict the score at x = 10.
At x = 1, ln 1 = 0, so a = 4.
At x = 5: 9 = 4 + b ln 5, so b = 5 / ln 5 ≈ 5 / 1.6094 ≈ 3.11. The model is y ≈ 4 + 3.11 ln x.
At x = 10: y ≈ 4 + 3.11(2.3026) ≈ 11.2. Since 10 is outside the given x-values of 1 to 5, this is an extrapolation and should be treated with caution.
y ≈ 4 + 3.11 ln x; y(10) ≈ 11.2
1. Ground motion y is 10x for a quake of magnitude x. What magnitude has ground motion of 5,000, to the nearest hundredth?
Explanation: Solve 10x = 5000: x = log 5000 ≈ 3.70. Magnitude is the logarithm of the ground motion.
2. A regression gives y = 5 + 2 ln x. What is the predicted y at x = 4, to the nearest hundredth?
Explanation: y = 5 + 2 ln 4 = 5 + 2(1.3863) = 7.77.
3. For y = 3 + 2 ln x, what value of x gives y = 9?
Explanation: Solve 3 + 2 ln x = 9: 2 ln x = 6, ln x = 3, so x = e3 ≈ 20.09.
4. A model is built from data with x-values from 1 to 20. Predicting at x = 12 is an example of what?
Explanation: 12 lies inside the range of the data, so the prediction is an interpolation. A prediction at x = 35 would be an extrapolation.
On a semi-log plot the x-axis is ordinary, but the y-axis is logarithmic: the spacing of tick marks depends on log y, so each step up multiplies the value instead of adding to it. This lets you display data that cover a huge range of values. The key fact is that an exponential function becomes a straight line on a semi-log plot. To see why, take a log of both sides of y = a·bx: log y = x·log b + log a. This is a line with slope log b and intercept log a. Similarly, for y = a·ekx, ln y = kx + ln a. This process is called linearizing. If data points fall on a line on a semi-log plot, an exponential model fits well, and the line’s slope and intercept give you b and a.
Worked Example: Linearize y = 3·2x, and decide how to read the model from the line.
Take log of both sides: log y = log 3 + x log 2.
So on a semi-log plot the line has slope log 2 ≈ 0.301 and intercept log 3 ≈ 0.477.
Going the other way, if a semi-log line has slope 0.301 and intercept 0.477, then b = 100.301 ≈ 2 and a = 100.477 ≈ 3.
Half-life version: a = a0(1/2)t/n becomes log2 a = log2 a0 − t/n, so the slope is −1/n. A slope of −0.05 per second means n = 20 seconds.
log y = x log 2 + log 3
1. On a semi-log plot, the graph of y = a·bx appears as:
Explanation: Taking the log gives log y = (log b)x + log a, which is linear in x with slope log b and intercept log a.
2. What are the slope and intercept of the linearized form of y = 4·10x, using log y?
Explanation: log y = log 4 + x log 10 = x + log 4. The slope is log 10 = 1 and the intercept is log 4 ≈ 0.602.
3. A semi-log analysis gives ln y = 0.3x + 2. Which exponential model matches?
Explanation: Exponentiate both sides: y = e0.3x+2 = e2·e0.3x ≈ 7.39e0.3x. The intercept is ln a, not a.
4. For a decay a = a0(1/2)t/n, the linearized line log2 a = log2 a0 − t/n has slope −0.05 per second. What is the half-life n?
Explanation: The slope is −1/n = −0.05, so n = 1/0.05 = 20 seconds.
log without a base means base 10, and ln means base e ≈ 2.718. Both obey the same laws. Use ln when the problem involves e, such as continuous growth, and log when it involves powers of 10.
A positive base raised to any real power is always positive, so no exponent can produce 0 or a negative number. That is why the domain of y = logb x is x > 0.
Exponential data multiply by a constant factor over equal steps and curve upward, logarithmic data rise quickly and then flatten, and linear data change by a constant amount. A scatterplot and the residual plot of each candidate help you choose.
The Exponential and Logarithmic Functions quiz in the AP Precalculus QBank gives you more practice on this topic, and you can browse every other topic in the full AP Precalculus QBank.