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Graphs of logarithms

Domain, asymptote, points · Algebra 2 · 20 problems, easy to harder

▶ Watch the first one solved

How we do it

  1. Find the domain
  2. Vertical asymptote
  3. Find some points
  4. Draw the graph
  5. Check

The steps for Graph y = log_2 x. Every solution names its steps, so you always know what comes next.

Worked examples

Graph y = log_2 x Easy

The question

Question: graph y = log₂ x.

y = log2 x

A log is a power.

log₂ of c is the power of 2 that makes c.

log₂ c = the power of 2 that gives c

1 Find the domain

2 to any power is always more than 0.

So a log only takes positive numbers.

The inside must be more than 0.

a log only takes numbers > 0

The inside is x.

So x > 0.

So the domain is x > 0.

x > 0

Books also write it as (0, ∞).

The round bracket means 0 itself is not in.

∞ means it goes on for ever.

interval: (0, ∞)

2 Vertical asymptote

At the edge, the inside is 0: x = 0.

Close to it, the graph goes down and down.

This is the vertical asymptote.

x = 0 → x = 0

Careful!

There is no log of 0, and no log of a negative number.

No power of 2 gives them.

Careful: log of 0 or a negative: no value

3 Find some points

The log is easy when the inside is a power of 2: 1/2, 1, 2 and 4.

We choose x to make that happen.

inside = 12, 1, 2, 4

Make the inside 1/2: log₂ of 1/2 is −1, because 2 to the power −1 is 1/2.

So y is −1.

x = 12 → x = 12, y = log2 12 = −1

Inside 1: log₂ of 1 is 0, because 2 to the power 0 is 1.

So y is 0.

x = 1 → x = 1, y = log2 1 = 0

Inside 2: log₂ of 2 is 1, because 2 to the power 1 is 2.

So y is 1.

x = 2 → x = 2, y = log2 2 = 1

Inside 4: log₂ of 4 is 2, because 2 to the power 2 is 4.

So y is 2.

x = 4 → x = 4, y = log2 4 = 2

4 Draw the graph

Draw the grid.

The asymptote x = 0, dashed.

Plot the points.

Join them with a smooth curve.

Near x = 0 it goes down along the dashed line, but never touches it.

So: domain x > 0, asymptote x = 0.

= domain x > 0, asymptote x = 0

5 Check

Let's check: at x = 0 the inside is 0, and no power of 2 is 0.

So there is no point there: the asymptote is right.

x = 0: inside = 0, log2 0: no power of 2 is 0

Let's look back.

First: find the domain.

Then: vertical asymptote.

Then: find some points.

Then: draw the graph.

Last: check.

Answer
= domain x > 0, asymptote x = 0

▶ Watch it solved

Graph y = log_3(x) + 1 Medium

The question

Question: graph y = log₃ x + 1.

y = log3 x + 1

A log is a power.

log₃ of c is the power of 3 that makes c.

log₃ c = the power of 3 that gives c

1 Find the domain

3 to any power is always more than 0.

So a log only takes positive numbers.

The inside must be more than 0.

a log only takes numbers > 0

The inside is x.

So x > 0.

So the domain is x > 0.

x > 0

Books also write it as (0, ∞).

The round bracket means 0 itself is not in.

∞ means it goes on for ever.

interval: (0, ∞)

2 Vertical asymptote

At the edge, the inside is 0: x = 0.

Close to it, the graph goes down and down.

This is the vertical asymptote.

x = 0 → x = 0

Careful!

There is no log of 0, and no log of a negative number.

No power of 3 gives them.

Careful: log of 0 or a negative: no value

3 Find some points

The log is easy when the inside is a power of 3: 1/3, 1, 3 and 9.

We choose x to make that happen.

inside = 13, 1, 3, 9

Make the inside 1/3: log₃ of 1/3 is −1, because 3 to the power −1 is 1/3.

So y is 0.

x = 13 → x = 13, y = log3 13 = −1 + 1 = 0

Inside 1: log₃ of 1 is 0, because 3 to the power 0 is 1.

So y is 1.

x = 1 → x = 1, y = log3 1 = 0 + 1 = 1

Inside 3: log₃ of 3 is 1, because 3 to the power 1 is 3.

So y is 2.

x = 3 → x = 3, y = log3 3 = 1 + 1 = 2

Inside 9: log₃ of 9 is 2, because 3 to the power 2 is 9.

So y is 3.

x = 9 → x = 9, y = log3 9 = 2 + 1 = 3

4 Draw the graph

Draw the grid.

The asymptote x = 0, dashed.

Plot the points.

Join them with a smooth curve.

Near x = 0 it goes down along the dashed line, but never touches it.

So: domain x > 0, asymptote x = 0.

= domain x > 0, asymptote x = 0

5 Check

Let's check: at x = 0 the inside is 0, and no power of 3 is 0.

So there is no point there: the asymptote is right.

x = 0: inside = 0, log3 0: no power of 3 is 0

The purple curve is y = log₃ x, the plain log.

Ours is that curve moved up 1.

The same shape.

It works!

Let's look back.

First: find the domain.

Then: vertical asymptote.

Then: find some points.

Then: draw the graph.

Last: check.

Answer
= domain x > 0, asymptote x = 0

▶ Watch it solved

Graph f(x) = log_2(x + 3) - 2 Harder

The question

Question: graph y = log₂(x + 3) − 2.

y = log2(x + 3) − 2

A log is a power.

log₂ of c is the power of 2 that makes c.

log₂ c = the power of 2 that gives c

1 Find the domain

2 to any power is always more than 0.

So a log only takes positive numbers.

The inside must be more than 0.

a log only takes numbers > 0

The inside is x + 3.

So x + 3 > 0.

x + 3 > 0

To remove the + 3, we take away 3 on both sides.

x + 3 − 3 > 0 − 3

So the domain is x > −3.

x > −3

Books also write it as (−3, ∞).

The round bracket means −3 itself is not in.

∞ means it goes on for ever.

interval: (−3, ∞)

2 Vertical asymptote

At the edge, the inside is 0: x = −3.

Close to it, the graph goes down and down.

This is the vertical asymptote.

x + 3 = 0 → x = −3

Careful!

The asymptote is where the inside is 0, not where x is 0.

Here that is x = −3.

Careful: not asymptote x = 0 ✗

3 Find some points

The log is easy when the inside is a power of 2: 1/2, 1, 2 and 4.

We choose x to make that happen.

inside = 12, 1, 2, 4

Make the inside 1/2: x is −5/2.

log₂ of 1/2 is −1, because 2 to the power −1 is 1/2.

So y is −3.

x + 3 = 12 → x = −52, y = log2 12 = −1 − 2 = −3

Inside 1: x is −2.

log₂ of 1 is 0, because 2 to the power 0 is 1.

So y is −2.

x + 3 = 1 → x = −2, y = log2 1 = 0 − 2 = −2

Inside 2: x is −1.

log₂ of 2 is 1, because 2 to the power 1 is 2.

So y is −1.

x + 3 = 2 → x = −1, y = log2 2 = 1 − 2 = −1

Inside 4: x is 1.

log₂ of 4 is 2, because 2 to the power 2 is 4.

So y is 0.

x + 3 = 4 → x = 1, y = log2 4 = 2 − 2 = 0

4 Draw the graph

Draw the grid.

The asymptote x = −3, dashed.

Plot the points.

Join them with a smooth curve.

Near x = −3 it goes down along the dashed line, but never touches it.

So: domain x > −3, asymptote x = −3.

= domain x > −3, asymptote x = −3

5 Check

Let's check: at x = −3 the inside is 0, and no power of 2 is 0.

So there is no point there: the asymptote is right.

x = −3: inside = 0, log2 0: no power of 2 is 0

The purple curve is y = log₂ x, the plain log.

Ours is that curve moved left 3 and down 2.

The same shape.

It works!

Let's look back.

First: find the domain.

Then: vertical asymptote.

Then: find some points.

Then: draw the graph.

Last: check.

Answer
= domain x > −3, asymptote x = −3

▶ Watch it solved

Practice

Try each one, then open the answer. Stuck? Open the problem to watch every step.

Easy

  1. graph y = log_2 x
    Answer
    = domain x > 0, asymptote x = 0
  2. graph y = log_2(x - 1)
    Answer
    = domain x > 1, asymptote x = 1
  3. graph y = log_2(x + 1)
    Answer
    = domain x > −1, asymptote x = −1
  4. graph y = log_3 x
    Answer
    = domain x > 0, asymptote x = 0
  5. graph y = log_2(x) + 1
    Answer
    = domain x > 0, asymptote x = 0
  6. graph y = log_2(x) - 2
    Answer
    = domain x > 0, asymptote x = 0
  7. graph y = log x
    Answer
    = domain x > 0, asymptote x = 0
  8. graph y = log_3(x - 2)
    Answer
    = domain x > 2, asymptote x = 2

Medium

  1. graph y = log_3(x) + 1
    Answer
    = domain x > 0, asymptote x = 0
  2. graph y = log(x + 2)
    Answer
    = domain x > −2, asymptote x = −2
  3. graph y = -log_2 x
    Answer
    = domain x > 0, asymptote x = 0
  4. graph y = log_2(x - 2) + 1
    Answer
    = domain x > 2, asymptote x = 2
  5. graph y = log_3(x + 1) - 1
    Answer
    = domain x > −1, asymptote x = −1
  6. graph y = -log_3(x)
    Answer
    = domain x > 0, asymptote x = 0

Harder

  1. graph f(x) = log_2(x + 3) - 2
    Answer
    = domain x > −3, asymptote x = −3
  2. graph y = -log_2(x - 1)
    Answer
    = domain x > 1, asymptote x = 1
  3. graph y = -log_2(x + 2) + 1
    Answer
    = domain x > −2, asymptote x = −2
  4. graph y = log(x - 1)
    Answer
    = domain x > 1, asymptote x = 1
  5. graph y = log(x) + 2
    Answer
    = domain x > 0, asymptote x = 0
  6. graph y = -log(x)
    Answer
    = domain x > 0, asymptote x = 0