Exponential & Logarithmic Functions: Test Review

Everything for the Unit 3 test on one page: the key facts from all 4 lessons, the mistakes that cost the most marks, and a practice test that points you back to the lessons you need.

Key facts

Introduction to Logarithms

Definition: $lo g_{b} x = y ⟺ b^{y} = x$ , with $b > 0$ , $b \neq = 1$ , $x > 0$ .
A logarithm is an exponent — the answer to $lo g_{b} x$ is the power you put on $b$ to make $x$ .
$lo g_{b} 1 = 0$ and $lo g_{b} b = 1$ for every valid base $b$ .
$lo g x$ with no base written means $lo g_{10} x$ , the common logarithm.
$lo g_{b} 0$ and $lo g_{b} (negative)$ are undefined, because $b^{y} > 0$ for every real $y$ .
$b^{l o g_{b} x} = x$ and $lo g_{b} b^{n} = n$ — the two operations undo each other.

Laws of Logarithms

Product law: $lo g_{b} (mn) = lo g_{b} m + lo g_{b} n$ .
Quotient law: $lo g_{b} (\frac{m}{n}) = lo g_{b} m - lo g_{b} n .$
Power law: $lo g_{b} (m^{n}) = n lo g_{b} m$ — this is the one that frees an exponent.
Change of base: $lo g_{b} x = \frac{lo g _{a} x}{lo g _{a} b}$ , usually with $a = 10$ .
There is no law for $lo g_{b} (m + n)$ or $lo g_{b} (m - n)$ — a logarithm of a sum cannot be split.
All three laws require a single common base; $lo g_{2} 5 + lo g_{3} 5$ cannot be combined.

Graphs of Logarithmic Functions

$y = lo g_{b} x$ is the inverse of $y = b^{x}$ ; its graph is the reflection of $y = b^{x}$ in the line $y = x$ .
Parent function: domain $x > 0$ , range $y \in R$ , vertical asymptote $x = 0$ , $x$ -intercept $(1, 0)$ , and it passes through $(b, 1)$ .
Transformed form: $y = a lo g_{b} (k (x - d)) + c$ — same parameter roles as every other function family.
The vertical asymptote moves to $x = d$ ; the vertical shift $c$ does not move it.
Domain of a transformed logarithm: solve $k (x - d) > 0$ . It is $x > d$ when $k > 0$ and $x < d$ when $k < 0$ .
For $b > 1$ , $y = lo g_{b} x$ has no maximum and no horizontal asymptote — it rises forever, just very slowly.

Solving Exponential and Logarithmic Equations

To free a variable from an exponent, take $lo g$ of both sides and apply the power law: $b^{x} = M \Rightarrow x = \frac{lo g M}{lo g b}$ .
If both sides can be written with a common base, match bases instead — it is faster and gives an exact answer.
$lo g_{b} A = lo g_{b} B \Rightarrow A = B$ ; and $lo g_{b} A = c \Rightarrow A = b^{c}$ .
State the domain before solving: every argument must be positive. Every candidate root must then be tested against it, and extraneous roots rejected explicitly.
Log scales: $pH = - lo g [H^{+}]$ , $β = 10 lo g (\frac{I}{I _{0}})$ with $I_{0} = 1 0^{- 12}$ W/m², and $M = lo g (\frac{I}{I _{0}})$ .
A difference of $n$ units on a Richter or pH scale means a factor of $1 0^{n}$ ; a difference of $n$ decibels means a factor of $1 0^{n /10}$ .

Where marks get lost

Treating $lo g_{2} 8$ as some kind of multiplication of 2 and 8. It is a question — "2 to what power gives 8?" — whose answer is 3. (Introduction to Logarithms)
Swapping the base and the argument, e.g. writing $lo g_{2} 8 = lo g_{8} 2$ . The small subscript is always the base, the big number is always the argument, and the two give completely different answers ( $3$ versus $\frac{1}{3}$ ). (Introduction to Logarithms)
Claiming $lo g_{b} 0 = 0$ . Nothing raised to any power gives zero, so this logarithm is undefined; it is $lo g_{b} 1$ that equals 0. (Introduction to Logarithms)
Trying to evaluate a logarithm of a negative number such as $lo g (- 100)$ — the argument must be positive. (Introduction to Logarithms)
Forgetting that a missing base means base 10, and reading $lo g 1000$ as if the base were unknown or arbitrary. (Introduction to Logarithms)
Writing $lo g (x + y) = lo g x + lo g y$ . The product law splits a product, never a sum. Test it: $lo g_{2} (4 + 4) = 3$ but $lo g_{2} 4 + lo g_{2} 4 = 4$ . (Laws of Logarithms)
Writing $\frac{lo g m}{lo g n} = lo g m - lo g n$ . The quotient law needs the division to be inside the logarithm, as in $lo g \frac{m}{n}$ . A quotient of two logarithms is a change of base, not a difference. (Laws of Logarithms)
Confusing $(lo g x)^{2}$ with $lo g x^{2}$ . Only the second one equals $2 lo g x$ ; the first is the logarithm multiplied by itself. (Laws of Logarithms)
Applying the power law to a coefficient that is not an exponent: $lo g_{b} (3 x^{4}) = lo g_{b} 3 + 4 lo g_{b} x,$ not $4 lo g_{b} (3 x)$ . The exponent 4 belongs to $x$ alone. (Laws of Logarithms)
Combining logarithms with different bases, such as trying to add $lo g_{2} 5$ and $lo g_{3} 5$ with the product law. Change both to a common base first. (Laws of Logarithms)
Giving the asymptote of $y = lo g_{2} (x - 5)$ as $y = 5$ or $y = 0$ . The asymptote of a logarithmic graph is vertical, so it must be written as $x = 5$ . (Graphs of Logarithmic Functions)
Writing the domain as $x > 0$ for every logarithm, even after a translation. Always solve "argument $> 0$ " for the actual function in front of you. (Graphs of Logarithmic Functions)
Assuming the graph always sits to the right of its asymptote. When $k < 0$ the graph is reflected in the $y$ -axis and sits to the left, as in $y = lo g (3 - x)$ . (Graphs of Logarithmic Functions)
Reading $d$ off an unfactored argument: in $y = lo g (2 x + 6)$ the asymptote is not $x = - 6$ . Factor first — $2 x + 6 = 2 (x + 3)$ — so the asymptote is $x = - 3$ . (Graphs of Logarithmic Functions)
Thinking the curve levels off or has a maximum because it looks flat on a calculator screen. It keeps climbing forever; $lo g_{10} x$ reaches 6 only when $x$ is a million. (Graphs of Logarithmic Functions)
Skipping the domain check on a logarithmic equation and reporting an extraneous root as a solution. Every candidate must be tested against the original arguments, and rejected roots must be named. (Solving Exponential and Logarithmic Equations)
Automatically rejecting any negative candidate. What has to be positive is the argument, not $x$ — in $lo g (x + 9) = 1$ the root $x = 1$ works, and a negative root would too if it kept $x + 9$ positive. (Solving Exponential and Logarithmic Equations)
Taking the logarithm of individual terms in a sum. From $3^{x} + 5 = 20$ you cannot write $x lo g 3 + lo g 5 = lo g 20$ ; first isolate the power as $3^{x} = 15$ , then take logarithms of both whole sides. (Solving Exponential and Logarithmic Equations)
"Cancelling" a logarithm instead of converting: from $lo g_{3} (x - 1) = 2$ , writing $x - 1 = 2$ rather than $x - 1 = 3^{2} = 9$ . (Solving Exponential and Logarithmic Equations)
Using $r$ instead of $1 + r$ as the base in a growth model, or mismatching the time units of $t$ and the rate (a monthly $t$ with an annual rate). (Solving Exponential and Logarithmic Equations)

Practice test

24 questions from across the unit. Each one shows the lesson it comes from.

Introduction to Logarithms
1
Evaluate $lo g_{3} 81$ .
Introduction to Logarithms
2
Rewrite $5^{3} = 125$ in logarithmic form.
Introduction to Logarithms
3
Evaluate $lo g_{9} 3$ . Give your answer as a decimal.
Introduction to Logarithms
4
Why is $lo g_{2} (- 8)$ undefined?
Introduction to Logarithms
5
Match each logarithm to its exact value.
$lo g_{6} 216$
$lo g_{4} 1$
$lo g_{10} 10000$
$lo g_{5} \frac{1}{25}$
$lo g_{27} 9$
Introduction to Logarithms
6
Rewrite $lo g_{2} 128 = 7$ in exponential form.
Laws of Logarithms
7
Evaluate $lo g 4 + lo g 25$ .
Laws of Logarithms
8
Evaluate $lo g_{6} 4 + lo g_{6} 9$ .
Laws of Logarithms
9
Expand $lo g_{b} (x^{2} y)$ fully.
Laws of Logarithms
10
Which of these is NOT a valid logarithm law?
Laws of Logarithms
11
Simplify $\frac{lo g _{5} 125}{lo g _{5} 5}$ .
Laws of Logarithms
12
Order the steps to condense $2 lo g x - \frac{1}{2} lo g y$ into one logarithm.
1. 1Rewrite the fractional exponent as a root: log x² − log √y
2. 2Apply the power law to each term: $lo g x^{2} - lo g y^{1/2}$
3. 3Apply the quotient law because the terms are subtracted: log(x²/√y)
Graphs of Logarithmic Functions
13
State the equation of the asymptote of $y = lo g_{2} (x - 5)$ .
Graphs of Logarithmic Functions
14
Find the value of $y$ on the graph of $y = lo g_{2} x$ when $x = 32$ .
Graphs of Logarithmic Functions
15
State the domain of $y = lo g (4 - x)$ .
Graphs of Logarithmic Functions
16
For $y = lo g_{2} (x - 3) + 4$ , find $y$ when $x = 11$ .
Graphs of Logarithmic Functions
17
State the asymptote of $y = lo g_{5} (3 x + 12)$ .
Graphs of Logarithmic Functions
18
Order the steps to find the domain and asymptote of $y = - 2 lo g_{3} (6 - 3 x) + 1$ .
1. 1Read off the parameters: k = −3 and d = 2
2. 2Solve 6 − 3x > 0 to get the domain x < 2
3. 3State the asymptote as x = d, so x = 2
4. 4Factor the coefficient of x out of the argument: 6 − 3x = −3(x − 2)
Solving Exponential and Logarithmic Equations
19
Solve $2^{x} = 50$ , correct to three decimal places.
Solving Exponential and Logarithmic Equations
20
Solve $9^{x} = 27$ by matching bases. Give $x$ as a fraction.
Solving Exponential and Logarithmic Equations
21
A sound has intensity $3.0 \times 1 0^{- 4}$ W/m². Find its sound level in decibels, to one decimal place. Use $I_{0} = 1 0^{- 12}$ W/m².
dB
Solving Exponential and Logarithmic Equations
22
Solve $5^{x} = 3$ , correct to three decimal places.
Solving Exponential and Logarithmic Equations
23
When solving $lo g (x + 2) + lo g (x - 1) = 1$ , the algebra produces $x = 3$ and $x = - 4$ . Which roots are acceptable?
Solving Exponential and Logarithmic Equations
24
Solve $lo g_{4} (x - 3) = 2$ .

Endless drill

Fresh numbers every time, for the calculation questions in this unit.

Endless drillnew numbers every time

Generating a problem…