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1. What is a logarithm? Logarithms as exponents

Warm-up

Three quick questions from earlier lessons.

Discover

Here is part of a table of powers of 10: $10^0 = 1$ , $10^1 = 10$ , $10^2 = 100$ , $10^3 = 1000$ .

Solve $10^x = 1000$ . Now try $10^x = 50$ : between which two whole numbers is $x$ ?

Key idea

A logarithm is an exponent. $\log_b x$ answers the question: " $b$ to what power gives $x$ ?"

\log_b x = y \iff b^y = x

(with

b > 0

b \ne 1

x > 0

)

Written without a base, $\log x$ means $\log_{10} x$ . So the solution of $10^x = 50$ is $x = \log 50 \approx 1.7$ .

Worked example

Evaluate $\log_2 32$ .

2^{y} = 32

Ask: 2 to what power gives 32?

y = 5

2⁵ = 32.

\log_2 32 = 5

So the logarithm is 5.

Your turn, guided

Your turn: evaluate $\log_3 81$ . Fill in the missing line.

3^{y} = 81

Ask: 3 to what power gives 81?

\log_3 81 = 4

3⁴ = 81.

Practice ★

Evaluate $\log_{ 4 } 16$ .
Evaluate $\log_{ 5 } 25$ .
Evaluate $\log_{ 4 } 256$ .
Evaluate $\log_{ 4 } 64$ .
Evaluate $\log_{ 2 } 8$ .

Watch out!

Practice ★★

Match each logarithm statement with the same statement written as a power.
1. $\log_{ 8 } y = 2$
2. $\log_{ 8 } y = 3$
3. $\log_{ 8 } y = -1$
4. $\log_{ 8 } y = \tfrac{1}{2}$
1. $y = 8^{2}$
2. $y = 8^{3}$
3. $y = \dfrac{1}{ 8 }$
4. $y = \sqrt{ 8 }$
Match each logarithm statement with the same statement written as a power.
1. $\log_{ 3 } y = 2$
2. $\log_{ 3 } y = 3$
3. $\log_{ 3 } y = -1$
4. $\log_{ 3 } y = \tfrac{1}{2}$
1. $y = 3^{2}$
2. $y = 3^{3}$
3. $y = \dfrac{1}{ 3 }$
4. $y = \sqrt{ 3 }$

Solve $6^x = 21$ . Write the exact answer using a logarithm.
Solve $7^x = 16$ . Write the exact answer using a logarithm.
Solve $2^x = 31$ . Write the exact answer using a logarithm.
Solve $7^x = 15$ . Write the exact answer using a logarithm.
Solve $2^x = 37$ . Write the exact answer using a logarithm.

Exit ticket

Three questions to check you have got it.

Solve $6^x = 21$ . Write the exact answer using a logarithm.
Solve $7^x = 16$ . Write the exact answer using a logarithm.
Solve $2^x = 31$ . Write the exact answer using a logarithm.

Summary card

Logarithms are exponents

$\log_b x = y$ means $b^y = x$ .
$\log x$ means $\log_{10} x$ .
The solution of $b^x = c$ is $x = \log_b c$ .

Watch out: $\log_b x$ is never $x \div b$ .

2. Asymptotes and shifts of exponential and logarithmic graphs

Warm-up

Three quick questions from earlier lessons.

Discover

Move the sliders and watch the graph of $y = a \cdot 2^x + k$ .

The graph of y = a·2^x + k, which changes as you move the sliders for a and k.

What does $k$ change?

It moves the graph and its horizontal asymptote up or down
It only moves the point where the graph crosses the y-axis
It makes the graph steeper

Key idea

Shifts move the asymptotes with the graph.

y = a \cdot b^{x} + k

has the horizontal asymptote

y = k

y = \log_b(x - h) + k

has the vertical asymptote

x = h

and domain

x > h

A vertical stretch by $a$ does not move a horizontal asymptote.

Worked example

Describe $y = 3 \cdot 2^x + 4$ .

Horizontal asymptote:

y = 4

As x → −∞, 3 · 2^x → 0.

y

-intercept:

3 + 4 = 7

At x = 0, 2⁰ = 1.

Your turn, guided

Your turn: $y = \log_3(x - 2) + 5$ . Fill in the missing line.

Domain:

x > 2

The argument x − 2 must be positive.

Vertical asymptote:

x = 2

The asymptote is where the argument equals 0.

Practice ★

The graph of $y = 3\cdot 4^x + 7$ has a horizontal asymptote $y = c$ . What is $c$ ?
The graph of $y = 8\cdot 5^x + 5$ has a horizontal asymptote $y = c$ . What is $c$ ?
The graph of $y = 2\cdot 3^x + 9$ has a horizontal asymptote $y = c$ . What is $c$ ?
The graph of $y = 6\cdot 5^x + 2$ has a horizontal asymptote $y = c$ . What is $c$ ?
The graph of $y = 7\cdot 2^x + 2$ has a horizontal asymptote $y = c$ . What is $c$ ?

Watch out!

Practice ★★

Draw the graph of $y = 1 \cdot 2^x + 3$ : place two points and the asymptote.
Draw the graph of $y = 1 \cdot 2^x + 1$ : place two points and the asymptote.

The graph of $y = \log_{ 4 }(x - 6) + 1$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 2 }(x - 4) + 8$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 2 }(x - 9) + 3$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 4 }(x - 9) + 5$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 4 }(x - 5) + 3$ has a vertical asymptote $x = c$ . What is $c$ ?

Exit ticket

Three questions to check you have got it.

The graph of $y = \log_{ 4 }(x - 6) + 1$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 2 }(x - 4) + 8$ has a vertical asymptote $x = c$ . What is $c$ ?
The graph of $y = \log_{ 2 }(x - 9) + 3$ has a vertical asymptote $x = c$ . What is $c$ ?

Summary card

Asymptotes after shifts

$y = a \cdot b^{x} + k$ : horizontal asymptote $y = k$ .
$y = \log_b(x - h) + k$ : vertical asymptote $x = h$ , domain $x > h$ .

3. Exponential models: half-life and doubling time

Warm-up

Three quick questions from earlier lessons.

Discover

A doctor gives a patient 80 mg of a medicine. Every 6 hours, half of what is left leaves the body.

How much is left after 6 hours? After 12? After 24? Does the amount ever reach 0?

Key idea

The half-life $h$ is the time it takes for an amount to halve; the doubling time $d$ is the time it takes to double.

Half-life

h

A(t) = A_0 \left(\tfrac{1}{2}\right)^{\frac{t}{h}}

. Doubling time

d

P(t) = P_0 \cdot 2^{\frac{t}{d}}

The exponent $\frac{t}{h}$ counts how many half-lives have passed.

Worked example

A sample of 80 mg has a half-life of 5 years. How much is left after 15 years?

\dfrac{15}{5} = 3

half-lives

Count the half-lives.

A = 80 \left(\tfrac{1}{2}\right)^3

Halve three times.

A = 10

80 ÷ 8 = 10.

Your turn, guided

Your turn: 96 mg, half-life 2 hours. How much is left after 8 hours? Fill in the missing line.

A = 96 \left(\tfrac{1}{2}\right)^4

8 ÷ 2 = 4 half-lives.

A = 6

96 ÷ 16 = 6.

Practice ★

A sample of 64 mg of a radioactive substance has a half-life of 2 years. How many mg are left after 6 years?
A sample of 64 mg of a radioactive substance has a half-life of 3 years. How many mg are left after 9 years?
A sample of 96 mg of a radioactive substance has a half-life of 5 years. How many mg are left after 20 years?
A sample of 320 mg of a radioactive substance has a half-life of 10 years. How many mg are left after 40 years?
A sample of 320 mg of a radioactive substance has a half-life of 2 years. How many mg are left after 10 years?

Watch out!

Practice ★★

A sample of 200 mg has a half-life of 2 years. How much is left after 8 years? Round to 2 decimal places.
A sample of 100 mg has a half-life of 3 years. How much is left after 11 years? Round to 2 decimal places.
A sample of 400 mg has a half-life of 3 years. How much is left after 4 years? Round to 2 decimal places.
A sample of 200 mg has a half-life of 3 years. How much is left after 10 years? Round to 2 decimal places.
A sample of 100 mg has a half-life of 4 years. How much is left after 11 years? Round to 2 decimal places.

Exit ticket

Three questions to check you have got it.

A dose of 200 mg of a medicine has a half-life of 2 hours. Write the amount $A(t)$ , in mg, left after $t$ hours.
A dose of 50 mg of a medicine has a half-life of 3 hours. Write the amount $A(t)$ , in mg, left after $t$ hours.
A dose of 500 mg of a medicine has a half-life of 4 hours. Write the amount $A(t)$ , in mg, left after $t$ hours.

Summary card

Half-life and doubling time

Half-life $h$ : $A_0 \left(\tfrac{1}{2}\right)^{\frac{t}{h}}$ . Doubling time $d$ : $P_0 \cdot 2^{\frac{t}{d}}$ .
$\frac{t}{h}$ = number of half-lives.

Watch out: $n$ half-lives divide by $2^n$ , not by $2n$ .

4. Properties of logarithms: product, quotient and power rules

Warm-up

Three quick questions from earlier lessons.

Discover

Use the fact that $\log_2 x$ is the exponent of 2. Compare $\log_2 4 + \log_2 8$ with $\log_2 32$ .

Why are they equal? Hint: $4 \cdot 8 = 32$ , and $2^2 \cdot 2^3 = 2^{2+3}$ .

Key idea

Logarithms turn multiplication into addition, because exponents add when powers multiply.

\log_b(MN) = \log_b M + \log_b N

;

\log_b \dfrac{M}{N} = \log_b M - \log_b N

;

\log_b(M^p) = p \log_b M

These hold for $M > 0$ and $N > 0$ .

Worked example

Evaluate $\log_6 4 + \log_6 9$ without a calculator.

= \log_6(4 \cdot 9)

Product rule.

Why is the first line true?

The log of a product is the sum of the logs
You can add inside the logarithm
The logs multiply

= \log_6 36

4 · 9 = 36.

= 2

6² = 36.

Your turn, guided

Your turn: evaluate $2\log_3 6 - \log_3 4$ . Fill in the missing line.

= \log_3 36 - \log_3 4

Power rule: 2 log₃ 6 = log₃ 36.

= \log_3 9

Quotient rule: 36 ÷ 4 = 9.

= 2

3² = 9.

Practice ★

Which expression is equal to $\log_{ 8 }(2 \cdot 7)$ ?
1. $\log_{ 8 } 2 + \log_{ 8 } 7$
2. $\log_{ 8 } 2 \cdot \log_{ 8 } 7$
3. $\log_{ 8 }(2 + 7)$
4. $7 \log_{ 8 } 2$
Which expression is equal to $\log_{ 7 }(4 \cdot 9)$ ?
1. $\log_{ 7 } 4 + \log_{ 7 } 9$
2. $\log_{ 7 } 4 \cdot \log_{ 7 } 9$
3. $\log_{ 7 }(4 + 9)$
4. $9 \log_{ 7 } 4$
Which expression is equal to $\log_{ 7 }(5 \cdot 3)$ ?
1. $\log_{ 7 } 5 + \log_{ 7 } 3$
2. $\log_{ 7 } 5 \cdot \log_{ 7 } 3$
3. $\log_{ 7 }(5 + 3)$
4. $3 \log_{ 7 } 5$

Evaluate $\log_{ 10 } 20 + \log_{ 10 } 5$ without a calculator.
Evaluate $\log_{ 10 } 40 + \log_{ 10 } 25$ without a calculator.
Evaluate $\log_{ 10 } 50 + \log_{ 10 } 20$ without a calculator.
Evaluate $\log_{ 10 } 25 + \log_{ 10 } 4$ without a calculator.
Evaluate $\log_{ 6 } 12 + \log_{ 6 } 18$ without a calculator.

Watch out!

Practice ★★

Evaluate $2\log_{ 2 } 6 - \log_{ 2 } 9$ without a calculator.
Evaluate $2\log_{ 3 } 6 - \log_{ 3 } 4$ without a calculator.
Evaluate $2\log_{ 5 } 15 - \log_{ 5 } 9$ without a calculator.
Evaluate $2\log_{ 3 } 18 - \log_{ 3 } 12$ without a calculator.
Evaluate $2\log_{ 3 } 15 - \log_{ 3 } 75$ without a calculator.

Write $\log_{ 3 } x + 2\log_{ 3 } 4$ as a single logarithm.
Write $\log_{ 6 } x + 2\log_{ 6 } 5$ as a single logarithm.
Write $\log_{ 7 } x + 2\log_{ 7 } 4$ as a single logarithm.

Expand $\log_{ 8 }\left(\dfrac{x^{ 4 }}{y^{ 5 }}\right)$ : write it with logarithms of $x$ and $y$ only.
Expand $\log_{ 5 }\left(\dfrac{x^{ 5 }}{y^{ 3 }}\right)$ : write it with logarithms of $x$ and $y$ only.
Expand $\log_{ 8 }\left(\dfrac{x^{ 2 }}{y^{ 5 }}\right)$ : write it with logarithms of $x$ and $y$ only.

Exit ticket

Three questions to check you have got it.

Evaluate $2\log_{ 2 } 6 - \log_{ 2 } 9$ without a calculator.
Evaluate $2\log_{ 3 } 6 - \log_{ 3 } 4$ without a calculator.
Evaluate $2\log_{ 5 } 15 - \log_{ 5 } 9$ without a calculator.

Summary card

Properties of logarithms ( $M, N > 0$ )

Product: $\log_b(MN) = \log_b M + \log_b N$ .
Quotient: $\log_b \frac{M}{N} = \log_b M - \log_b N$ .
Power: $\log_b(M^p) = p \log_b M$ .

Watch out: there is no rule for $\log_b(M + N)$ .

5. Change of base for logarithms

Warm-up

Three quick questions from earlier lessons.

Discover

Most calculators only have $\ln$ and $\log$ (base 10). How could you find $\log_2 10$ with one?

Hint: if $y = \log_2 10$ , then $2^y = 10$ . Take $\ln$ of both sides.

Key idea

Any logarithm can be written with natural logarithms (or with base 10).

\log_a c = \dfrac{\ln c}{\ln a} = \dfrac{\log c}{\log a}

The base $a$ always goes in the denominator.

Worked example

Evaluate $\log_4 8$ exactly.

\log_4 8 = \dfrac{\ln 8}{\ln 4}

Change of base.

= \dfrac{3\ln 2}{2\ln 2}

8 = 2³ and 4 = 2², then the power rule.

= \dfrac{3}{2}

ln 2 cancels.

Your turn, guided

Your turn: evaluate $\log_9 27$ . Fill in the missing line.

\log_9 27 = \dfrac{\ln 27}{\ln 9}

Change of base.

= \dfrac{3\ln 3}{2\ln 3}

27 = 3³ and 9 = 3².

= \dfrac{3}{2}

ln 3 cancels.

Practice ★

Place $\log_{2} 6$ on the number line.
Place $\log_{2} 24$ on the number line.
Place $\log_{2} 21$ on the number line.

Complete the change of base: $\log_{ 4 } 20 = \dfrac{\ln 20 }{\square}$ . What goes in the box?
Complete the change of base: $\log_{ 8 } 24 = \dfrac{\ln 24 }{\square}$ . What goes in the box?
Complete the change of base: $\log_{ 4 } 16 = \dfrac{\ln 16 }{\square}$ . What goes in the box?
Complete the change of base: $\log_{ 7 } 4 = \dfrac{\ln 4 }{\square}$ . What goes in the box?
Complete the change of base: $\log_{ 4 } 9 = \dfrac{\ln 9 }{\square}$ . What goes in the box?

Watch out!

Practice ★★

Write $\log_{ 8 } 16$ as a fraction in lowest terms.
Write $\log_{ 8 } 2$ as a fraction in lowest terms.
Write $\log_{ 9 } 81$ as a fraction in lowest terms.
Write $\log_{ 27 } 3$ as a fraction in lowest terms.
Write $\log_{ 9 } 243$ as a fraction in lowest terms.

Exit ticket

Three questions to check you have got it.

Evaluate $\log_{ 8 } 16$ exactly. Hint: write 8 and 16 as powers of the same number.
Evaluate $\log_{ 8 } 4$ exactly. Hint: write 8 and 4 as powers of the same number.
Evaluate $\log_{ 27 } 9$ exactly. Hint: write 27 and 9 as powers of the same number.

Summary card

Change of base

$\log_a c = \dfrac{\ln c}{\ln a}$ : the base goes below.

Check: $\log_2 5$ must be between 2 and 3, because $2^2 = 4$ and $2^3 = 8$ .

6. Solve equations of the form e^(kx) = a

Warm-up

Three quick questions from earlier lessons.

Discover

A colony of bacteria doubles every hour, so after $t$ hours there are $e^{0.69t}$ times as many. When will there be 7 times as many? Try to estimate before reading on: between which two whole hours?

Key idea

To undo $e^x$ , use its inverse function: $\ln$ .

e^x = a \iff x = \ln a

, for

a > 0

Since $\ln(e^{x}) = x$ , taking $\ln$ of both sides "removes" the exponential.

Worked example

Solve $e^{2x} = 7$ .

2x = \ln 7

Apply ln to both sides: ln undoes e.

x = \dfrac{\ln 7}{2}

Divide both sides by 2.

Your turn, guided

Your turn: solve $e^{3x} = 5$ . Fill in the missing line.

3x = \ln 5

Apply ln to both sides.

x = \dfrac{\ln 5}{3}

Divide both sides by 3.

Practice ★

Solve $e^{ 5x } = 25$ .
Solve $e^{ 5x } = 7$ .
Solve $e^{ 5x } = 9$ .
Solve $e^{ 3x } = 15$ .
Solve $e^{ 5x } = 10$ .

Watch out!

Practice ★★

Solve $e^{ 4x + 4 } = 12$ .
Solve $e^{ 6x + 5 } = 37$ .
Solve $e^{ 2x + 7 } = 16$ .
Solve $e^{ 3x + 6 } = 27$ .
Solve $e^{ 6x + 7 } = 39$ .

Exit ticket

Three questions to check you have got it.

Solve $e^{ 4x + 4 } = 12$ .
Solve $e^{ 6x + 5 } = 37$ .
Solve $e^{ 2x + 7 } = 16$ .

Summary card

Solving $e^{kx} = a$ (with $a > 0$ )

Take $\ln$ of both sides: $kx = \ln a$ .
Divide by $k$ : $x = \dfrac{\ln a}{k}$ .

Watch out: $e$ is not a factor. Never divide by $e$ .

One more step is optional: a challenge. Try it, or finish the lesson now.

Challenge ★★★

Solve $3e^{ 4x } + 6 = 60$ .
Solve $5e^{ 2x } + 6 = 25$ .
Solve $4e^{ 2x } + 8 = 36$ .

7. Solve exponential equations in any base with logarithms

Warm-up

Three quick questions from earlier lessons.

Discover

You can solve $2^x = 8$ in your head. What about $2^x = 7$ ?

Which two whole numbers is the answer between? How could logarithms give it exactly?

Key idea

To solve $a \cdot b^x = c$ : isolate the power, then take $\ln$ of both sides and use the power rule.

b^x = k \iff x = \dfrac{\ln k}{\ln b}

(for

k > 0

)

Worked example

Solve $3 \cdot 2^x = 21$ .

2^x = 7

Divide both sides by 3 to isolate the power.

x \ln 2 = \ln 7

Take ln of both sides; power rule.

x = \dfrac{\ln 7}{\ln 2} \approx 2.81

Divide by ln 2.

Put the steps for solving $a \cdot b^x = c$ in order.

Use the power rule: $x \ln b = \ln \frac{c}{a}$
Divide by $\ln b$
Take $\ln$ of both sides
Divide both sides by $a$

Your turn, guided

Your turn: solve $5^x = 12$ . Fill in the missing line.

x \ln 5 = \ln 12

Take ln of both sides; power rule.

x = \dfrac{\ln 12}{\ln 5}

Divide by ln 5.

Practice ★

Solve $6^x = 35$ . Give the exact answer using natural logarithms.
Solve $2^x = 10$ . Give the exact answer using natural logarithms.
Solve $7^x = 36$ . Give the exact answer using natural logarithms.
Solve $4^x = 20$ . Give the exact answer using natural logarithms.
Solve $2^x = 17$ . Give the exact answer using natural logarithms.

Watch out!

Practice ★★

Solve $2\cdot 3^x = 48$ . Give the exact answer.
Solve $6\cdot 4^x = 90$ . Give the exact answer.
Solve $4\cdot 4^x = 30$ . Give the exact answer.
Solve $6\cdot 5^x = 50$ . Give the exact answer.
Solve $2\cdot 4^x = 72$ . Give the exact answer.

Solve $4\cdot 2^x = 65$ step by step.
Solve $3\cdot 5^x = 63$ step by step.

Exit ticket

Three questions to check you have got it.

Solve $2\cdot 3^x = 48$ . Give the exact answer.
Solve $6\cdot 4^x = 90$ . Give the exact answer.
Solve $4\cdot 4^x = 30$ . Give the exact answer.

Summary card

Solving $a \cdot b^x = c$

Isolate the power: $b^x = \dfrac{c}{a}$ .
Take $\ln$ : $x \ln b = \ln \dfrac{c}{a}$ .
Divide: $x = \dfrac{\ln(c/a)}{\ln b}$ .

Watch out: never multiply $a$ into the base.

8. Solve logarithmic equations and check the domain

Warm-up

Three quick questions from earlier lessons.

Discover

Solve $\log_2 x = 3$ by rewriting it as a power.

Then think: can $\ln x$ ever be defined for $x = -2$ ? What does that mean for solutions you find?

Key idea

To solve a logarithmic equation: combine into one logarithm, rewrite as a power (or compare arguments), solve, then check the domain.

\log_b(\text{expression}) = n \iff \text{expression} = b^{n}

, and every argument of a logarithm must be positive

Worked example

Solve $\ln x + \ln(x - 2) = \ln 15$ .

\ln\big(x(x - 2)\big) = \ln 15

Product rule.

x^2 - 2x - 15 = 0

Equal logarithms have equal arguments.

(x - 5)(x + 3) = 0

Factor.

x = 5

x = −3 makes ln x undefined: reject it.

Your turn, guided

Your turn: solve $\log_2(x - 3) = 4$ . Fill in the missing line.

x - 3 = 2^4 = 16

Rewrite as a power.

x = 19

Add 3. Check: 19 − 3 = 16 > 0.

Practice ★

Put the steps for solving $\ln x + \ln(x - 5) = \ln 6$ in order.
- Combine with the product rule: $\ln\big(x(x - 5)\big) = \ln 6$
- Equal logarithms have equal arguments: $x(x - 5) = 6$
- Solve the quadratic equation
- Keep only the values where $x > 5$
Put the steps for solving $\ln x + \ln(x - 4) = \ln 21$ in order.
- Combine with the product rule: $\ln\big(x(x - 4)\big) = \ln 21$
- Equal logarithms have equal arguments: $x(x - 4) = 21$
- Solve the quadratic equation
- Keep only the values where $x > 4$

Solve $\log_{ 5 }(x - 8) = 4$ .
Solve $\log_{ 5 }(x - 9) = 3$ .
Solve $\log_{ 3 }(x - 8) = 4$ .
Solve $\log_{ 2 }(x - 2) = 4$ .
Solve $\log_{ 2 }(x - 6) = 4$ .

Watch out!

Two learners solved $\ln x + \ln(x - 3) = \ln 4$ . Compare their work.

Solution A

x^2 - 3x - 4 = 0

Product rule, then compare the arguments.

(x - 4)(x + 1) = 0

Factor.

x = 4

x = -1

Both values solve the quadratic.

Solution B

x^2 - 3x - 4 = 0

Product rule, then compare the arguments.

(x - 4)(x + 1) = 0

Factor.

x = 4

ln(−1) is not defined, so x = −1 is rejected.

Solution A is correct
Solution B is correct
Both are correct

Practice ★★

Solve $\ln x + \ln(x - 1) = \ln 42$ . Give every solution.
Solve $\ln x + \ln(x - 2) = \ln 24$ . Give every solution.
Solve $\ln x + \ln(x - 3) = \ln 18$ . Give every solution.
Solve $\ln x + \ln(x - 4) = \ln 45$ . Give every solution.
Solve $\ln x + \ln(x - 1) = \ln 72$ . Give every solution.

Exit ticket

Three questions to check you have got it.

Solve $\ln x + \ln(x - 2) = \ln 3$ .
Solve $\ln x + \ln(x - 2) = \ln 48$ .
Solve $\ln x + \ln(x - 1) = \ln 56$ .

Summary card

Solving logarithmic equations

Combine into one logarithm (product, quotient, power rules).
Rewrite as a power, or compare arguments.
Solve, then check that every argument is positive.

Answers: Exponential and logarithmic functions

1. What is a logarithm? Logarithms as exponents

$\log_{ 8 } y = 2$ → $y = 8^{2}$ · $\log_{ 8 } y = 3$ → $y = 8^{3}$ · $\log_{ 8 } y = -1$ → $y = \dfrac{1}{ 8 }$ · $\log_{ 8 } y = \tfrac{1}{2}$ → $y = \sqrt{ 8 }$
$\log_{ 3 } y = 2$ → $y = 3^{2}$ · $\log_{ 3 } y = 3$ → $y = 3^{3}$ · $\log_{ 3 } y = -1$ → $y = \dfrac{1}{ 3 }$ · $\log_{ 3 } y = \tfrac{1}{2}$ → $y = \sqrt{ 3 }$

$\frac{\ln{\left(21 \right)}}{\ln{\left(6 \right)}}$
$\frac{\ln{\left(16 \right)}}{\ln{\left(7 \right)}}$
$\frac{\ln{\left(31 \right)}}{\ln{\left(2 \right)}}$
$\frac{\ln{\left(15 \right)}}{\ln{\left(7 \right)}}$
$\frac{\ln{\left(37 \right)}}{\ln{\left(2 \right)}}$

$\frac{\ln{\left(21 \right)}}{\ln{\left(6 \right)}}$
$\frac{\ln{\left(16 \right)}}{\ln{\left(7 \right)}}$
$\frac{\ln{\left(31 \right)}}{\ln{\left(2 \right)}}$

2. Asymptotes and shifts of exponential and logarithmic graphs

$y = 2^{x} + 3$
$y = 2^{x} + 1$

3. Exponential models: half-life and doubling time

$8$
$8$
$6$
$20$
$10$

$\frac{25}{2}$ mg
$\frac{25 \sqrt[3]{2}}{4}$ mg
$100 \cdot 2^{\frac{2}{3}}$ mg
$\frac{25 \cdot 2^{\frac{2}{3}}}{2}$ mg
$\frac{25 \sqrt[4]{2}}{2}$ mg

$200 \cdot 2^{- \frac{t}{2}}$
$50 \cdot 2^{- \frac{t}{3}}$
$500 \cdot 2^{- \frac{t}{4}}$

4. Properties of logarithms: product, quotient and power rules

$\log_{ 8 } 2 + \log_{ 8 } 7$
$\log_{ 7 } 4 + \log_{ 7 } 9$
$\log_{ 7 } 5 + \log_{ 7 } 3$

$\frac{\ln{\left(16 x \right)}}{\ln{\left(3 \right)}}$
$\frac{\ln{\left(25 x \right)}}{\ln{\left(6 \right)}}$
$\frac{\ln{\left(16 x \right)}}{\ln{\left(7 \right)}}$

$\frac{4 \ln{\left(x \right)}}{\ln{\left(8 \right)}} - \frac{5 \ln{\left(y \right)}}{\ln{\left(8 \right)}}$
$\frac{5 \ln{\left(x \right)}}{\ln{\left(5 \right)}} - \frac{3 \ln{\left(y \right)}}{\ln{\left(5 \right)}}$
$\frac{2 \ln{\left(x \right)}}{\ln{\left(8 \right)}} - \frac{5 \ln{\left(y \right)}}{\ln{\left(8 \right)}}$

5. Change of base for logarithms

$1 + \frac{\ln{\left(3 \right)}}{\ln{\left(2 \right)}}$
$\frac{\ln{\left(3 \right)}}{\ln{\left(2 \right)}} + 3$
$\frac{\ln{\left(21 \right)}}{\ln{\left(2 \right)}}$

$\ln{\left(4 \right)}$
$\ln{\left(8 \right)}$
$\ln{\left(4 \right)}$
$\ln{\left(7 \right)}$
$\ln{\left(4 \right)}$

$\frac{4}{3}$
$\frac{1}{3}$
$2$
$\frac{1}{3}$
$\frac{5}{2}$

$\frac{4}{3}$
$\frac{2}{3}$
$\frac{2}{3}$

6. Solve equations of the form e^(kx) = a

$\frac{\ln{\left(25 \right)}}{5}$
$\frac{\ln{\left(7 \right)}}{5}$
$\frac{\ln{\left(9 \right)}}{5}$
$\frac{\ln{\left(15 \right)}}{3}$
$\frac{\ln{\left(10 \right)}}{5}$

$-1 + \frac{\ln{\left(12 \right)}}{4}$
$- \frac{5}{6} + \frac{\ln{\left(37 \right)}}{6}$
$- \frac{7}{2} + \frac{\ln{\left(16 \right)}}{2}$
$-2 + \frac{\ln{\left(27 \right)}}{3}$
$- \frac{7}{6} + \frac{\ln{\left(39 \right)}}{6}$

$-1 + \frac{\ln{\left(12 \right)}}{4}$
$- \frac{5}{6} + \frac{\ln{\left(37 \right)}}{6}$
$- \frac{7}{2} + \frac{\ln{\left(16 \right)}}{2}$

$\frac{\ln{\left(18 \right)}}{4}$
$\frac{\ln{\left(\frac{19}{5} \right)}}{2}$
$\frac{\ln{\left(7 \right)}}{2}$

7. Solve exponential equations in any base with logarithms

$\frac{\ln{\left(35 \right)}}{\ln{\left(6 \right)}}$
$\frac{\ln{\left(10 \right)}}{\ln{\left(2 \right)}}$
$\frac{\ln{\left(36 \right)}}{\ln{\left(7 \right)}}$
$\frac{\ln{\left(20 \right)}}{\ln{\left(4 \right)}}$
$\frac{\ln{\left(17 \right)}}{\ln{\left(2 \right)}}$

$\frac{\ln{\left(24 \right)}}{\ln{\left(3 \right)}}$
$\frac{\ln{\left(15 \right)}}{\ln{\left(4 \right)}}$
$\frac{\ln{\left(\frac{15}{2} \right)}}{\ln{\left(4 \right)}}$
$\frac{\ln{\left(\frac{25}{3} \right)}}{\ln{\left(5 \right)}}$
$\frac{\ln{\left(36 \right)}}{\ln{\left(4 \right)}}$

$\frac{\ln{\left(\frac{65}{4} \right)}}{\ln{\left(2 \right)}}$
$\frac{\ln{\left(21 \right)}}{\ln{\left(5 \right)}}$

$\frac{\ln{\left(24 \right)}}{\ln{\left(3 \right)}}$
$\frac{\ln{\left(15 \right)}}{\ln{\left(4 \right)}}$
$\frac{\ln{\left(\frac{15}{2} \right)}}{\ln{\left(4 \right)}}$

8. Solve logarithmic equations and check the domain

1. Combine with the product rule: $\ln\big(x(x - 5)\big) = \ln 6$ 2. Equal logarithms have equal arguments: $x(x - 5) = 6$ 3. Solve the quadratic equation 4. Keep only the values where $x > 5$
1. Combine with the product rule: $\ln\big(x(x - 4)\big) = \ln 21$ 2. Equal logarithms have equal arguments: $x(x - 4) = 21$ 3. Solve the quadratic equation 4. Keep only the values where $x > 4$

$633$
$134$
$89$
$18$
$22$