Secondary 5 Logarithmic Functions Worksheet

All 7 questions

Each question targets one named concept from the sheet. Read them here, or print the PDF — it has working space under each one.

1. Q1Finding the Rule of a Logarithmic Function

   The curve of a logarithmic function $f(x)=a{\log }_2(x-h)$ has the vertical asymptote $x=3$ and passes through the point $(7,-4)$.

   1. Find its rule.
   2. Give the domain of $f$ and the zero of $f$.
   3. Evaluate $f(19)$.
2. Q2Graphing the Logarithmic Function

   Consider $f(x)={\log }_2(x+4)-3$.

   1. Give the equation of the vertical asymptote, the domain and the range.
   2. Find the zero and the $y$-intercept, then sketch the curve using also the points of abscissa $-3$ and $-2$.
   3. Is $f$ increasing or decreasing? Justify.
3. Q3Properties of Logarithmic Functions

   Consider $f(x)=-2{\log }_3(x+6)+4$.

   1. Give the domain and the equation of the asymptote.
   2. Find the zero of $f$.
   3. State the variation of $f$ and justify it from the parameters.
   4. Give the intervals on which $f$ is positive and negative.
4. Q4Solving Problems Involving Logarithmic Functions

   A climbing gym rates how dusty its holds are with a cleanliness index $$C(n)=40-12\log n,$$ where $n\ge 1$ is the number of climbers since the last cleaning.

   1. What is the index after $10$ climbers?
   2. After how many climbers does the index fall to $4$?
5. Q5The Inverse of the Logarithmic Function

   Find the rule of the inverse of $f(x)={\log }_3(x-2)+4$, then state the domain of $f$ and the domain of $f^{-1}$.

6. Q6The Logarithmic Function

   Consider $f(x)={\log }_{1/2}(x+3)-1$. State its domain, the equation of its asymptote and its direction of variation, then find its zero. Explain in one sentence why $f(-3)$ does not exist.

7. Q7The Role of the Parameters in a Logarithmic Function

   Consider $f(x)=2{\log }_3(x+4)-6$.

   1. Give the equation of the vertical asymptote and the domain.
   2. Is $f$ increasing or decreasing? Justify with the parameters.
   3. Find the zero of $f$.

The 6 challenge problems for this topic are a separate, paid sheet and are not reproduced here.

How to do every concept on this sheet

This is the part worksheet sites usually leave out. Below is the actual reasoning behind each group of questions — not a full solution set, the decisions that get you to one. Read it before you start, or after you get stuck.

A logarithm is an exponent — read every rule that way

Everything on this sheet becomes routine once you stop treating log as an unexplained button and read it as the question it actually asks:

log₂(16) is "what exponent turns 2 into 16?", so it is 4. That is not a calculation, it is a reading, and Q1(c), Q2 and Q4 are all designed to be answered by reading rather than by calculator. Two special cases are worth having permanently on hand: log_c(1) = 0 for every base, because c⁰ is always 1, and log_c(c) = 1. Those two are the reason a point of a logarithmic graph can drop out of a substitution with no work.

One notation convention: a logarithm written with no base at all, as in Q4's log n, means base 10. So log 1000 is 3.

The domain is the whole first half of most questions

A logarithm accepts only strictly positive arguments, because no exponent applied to a positive base ever produces zero or a negative number. That restriction is not a technicality bolted on afterwards; it is where the vertical asymptote comes from.

Solving: isolate the logarithm, then convert

Q1(b), Q2(b), Q3(b), Q4(b), Q6 and Q7(c) are all the same two-step method, and the order matters.

Step one: get the logarithm completely alone. Move the + k across, then divide by a. The equation must read log_c(argument) = number with nothing multiplying or added to the log.

Step two: rewrite it as an exponential statement using the definition above, then solve the ordinary equation that remains. For 4·log₅(x − 1) − 8 = 0:

Finish by checking the solution lies in the domain. A logarithmic equation can produce a value that satisfies the algebra but makes an argument negative, and such a value must be rejected explicitly, with the reason written down.

Finding the rule from an asymptote and a point (Q1)

The asymptote gives h immediately — it is the value that makes the argument zero, so for an asymptote at x = 3 the argument is x − 3. That leaves one unknown, and one point on the curve determines it.

Substitute the point, evaluate the logarithm as an exponent (this is where the "read it, don't compute it" habit pays), and solve a one-step equation. Then verify with a second value. If a question supplies two points instead of an asymptote, substitute both and divide one equation by the other to clear the coefficient — and be prepared for the domain to reject one of the resulting candidates for h.

Increasing or decreasing: the base and the coefficient vote together

Q2(c), Q3(c), Q6 and Q7(b) all ask you to justify the variation from the parameters, which means the answer is a sentence about signs, not a pair of substituted values.

k does not appear in that reasoning, and neither does h: shifting a curve sideways or vertically cannot change whether it rises. Q3's + 4 and Q7's − 6 move the zero, nothing more.

Sign intervals come free from monotonicity (Q3)

Because the function crosses the x-axis exactly once and never turns around, the domain splits into precisely two pieces at the zero: positive on one, negative on the other. You do not need a table of signs — you need the zero and the direction.

If the function is decreasing with its zero at x = z and its asymptote at x = p, then it is positive on ]p, z[ and negative on ]z, +∞[. Both intervals are open at p because the function does not exist there, and open at z because the function is zero there rather than positive or negative. If it is increasing, the two labels swap. That is also how a logarithmic inequality is solved: find where equality holds, then use the direction of variation to decide which side of it you keep, and intersect with the domain.

Sketching (Q2)

Draw the vertical asymptote as a dashed line first, then choose your points so the logarithm evaluates exactly. For base 2, pick x values that make the argument 1, 2, 4, 8 — the logarithms are then 0, 1, 2, 3 and the points are exact. That is precisely why Q2 names the abscissas −3 and −2: they make the argument 1 and 2.

The shape to aim for: plunging steeply along the asymptote, then flattening out as it moves away, but continuing to rise forever without any ceiling. Two errors to avoid: the curve must not stop or level off at a horizontal line — there is no horizontal asymptote — and it must not appear on both sides of the vertical one.

Contextual logarithms (Q4)

Logarithmic models show up where a quantity responds to multiplying the input rather than adding to it: sound levels, pH, magnitudes, and index scales like the one in Q4. The signature is that equal steps in the output correspond to equal factors in the input — going from 10 to 100 to 1000 climbers moves the index by the same amount each time.

Part (a) is direct substitution, and it is meant to be exact: with base 10, log 10 is 1. Part (b) reverses the model, which is the isolate-then-convert method again, ending with a power of 10. Finish by reading the answer back into the situation — a number of climbers is a whole number, and the stated condition n ≥ 1 is the model's domain, which exists because log 0 does not.

The inverse is an exponential (Q5)

Swap x and y, isolate the logarithm, then convert with the definition — the same two steps as solving, applied to a rule instead of a number. From x = log₃(y − 2) + 4 you get log₃(y − 2) = x − 4 and therefore y = 3^x−4 + 2.

What the parameter b does that nothing else can

In the full form a·log_c(b(x − h)) + k, the sign of b decides which side of the asymptote the curve lives on, because the argument has to stay positive. With b > 0 the domain runs to the right of x = h; with b < 0 it runs to the left, and the curve is the mirror image across the vertical line x = h. Its size, though, only translates the curve vertically, since multiplying the argument by a constant adds a constant to the logarithm. That asymmetry — sign matters, magnitude is absorbable — is unique to this family, and it is why the domain check in step 1 above must use the actual argument rather than an assumption about where the curve sits.

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Getting the most out of it

Write the domain before you write anything else

Make it a physical habit: on every question, the first line you put on the page is the argument set greater than zero, solved. It gives you the asymptote for free and it is the filter every later answer has to pass through.

Learn to read logarithms instead of computing them

Cover the answers and say aloud what log₂(32), log₃(81), log(10 000) and log₅(1) are. Every number in these questions is chosen to come out whole, so if you reach for a calculator you have probably misread the base or left a coefficient attached to the logarithm.

Redo the variation questions as one paragraph

Q2, Q3, Q6 and Q7 each ask for a justification. Once you have finished the sheet, write a single paragraph explaining how the base and the sign of a combine to fix the direction, then compare it against what you wrote in each question. Those are the marks that get lost to one-word answers.

Common questions