Logarithm Graphs: Follow the Input, Watch the Output

CBSE Mathematics · 2026–27 · Class 11

Learn with Rajesh Sir

A curve can answer several questions at once: which inputs are allowed, whether outputs rise or fall, and how close you can move to a boundary. Build a logarithm graph from a few exact points, then change its base and shape.

By the end, you can…

Identify the domain, range, intercept and asymptote of a logarithm graph.
Explain the two behaviours for bases above and below 1.
Match points on exponential and logarithmic inverse graphs.
Read shifts and reflections without losing the argument restriction.
Distinguish a finite graph window from the full function.

About 5 minutes

TOPIC 01Class 11 · functions and graphs

Start with three points and one boundary

For $f(x)=\log_b x$ , where $b>0$ and $b\ne1$ , choose arguments that are easy powers of the base. Three anchors are

(1,0),\qquad(b,1),\qquad(1/b,-1).

The first coordinate is the input; the second is the logarithm value. With base 2, these become $(1,0)$ , $(2,1)$ and $(1/2,-1)$ . Negative outputs are normal even though every input must be positive.

The domain is $(0,\infty)$ . The range is all real numbers: given any real output $y$ , the positive input $b^y$ produces it. A chosen codomain is a target set in a function’s definition; the range is the set of values actually attained.

The graph crosses the horizontal axis at $(1,0)$ . It has no vertical-axis intercept, because $x=0$ is not allowed. The line $x=0$ is a vertical asymptote, a boundary the curve approaches.

A screen can display only part of this graph. A plotted window ending at $x=10$ , or outputs between $-3$ and 3, does not shorten the mathematical domain or range.

Work it through

Worked example: build a base-3 point table. From $3^{-2}=1/9$ , $3^{-1}=1/3$ , $3^0=1$ , $3^1=3$ and $3^2=9$ , obtain

Input $x$	$1/9$	$1/3$	1	3	9
$\log_3 x$	−2	−1	0	1	2

Every point has a positive first coordinate. The table does not stop the function at $-2$ or 2; choosing further powers produces further outputs.

Enable JavaScript to explore the inputs. The explanation and worked example remain available here.

Pause. Think. Try.

1. Which point lies on

y=\log_b x

for every valid base

b

(0,1)

(1,0)

(0,0)

(1,1)

A hint, please

Which exponent gives 1 for every allowed base?

Read the answer and explanation

$(1,0)$

b^0=1

, so

\log_b1=0

. Thus

(1,0)

is always on the graph.

2. What are the domain and range of

y=\log_2 x

?Domain

\mathbb R

; range

(0,\infty)

.Domain

(0,\infty)

; range

(0,\infty)

.Domain

(0,\infty)

; range

\mathbb R

.Domain

[0,\infty)

; range

\mathbb R

A hint, please

Check which coordinate must be positive.

Read the answer and explanation

Domain $(0,\infty)$ ; range $\mathbb R$ .

Only the input must be positive. For any real

y

, choosing

x=2^y>0

gives that output, so the range is

\mathbb R

About 5 minutes

TOPIC 02Class 11 · functions and graphs

A base below 1 turns the graph around

Change the base without changing the input rule $x>0$ . Two different patterns appear. If $b>1$ , $\log_b x$ is strictly increasing: a larger positive input produces a larger output. If $0<b<1$ , it is strictly decreasing: a larger positive input produces a smaller output.

Both families pass through $(1,0)$ . For a base above 1, inputs between 0 and 1 give negative outputs, and inputs above 1 give positive outputs. For a base below 1, these signs reverse.

Near the boundary, the behaviours also reverse. As $x$ approaches zero from the positive side, $\log_2 x$ falls without bound, while $\log_{1/2}x$ rises without bound. Neither function has a value at zero. There is no “first positive input”: between zero and any positive input lies another positive number.

Watch two marked points as you change the base. The input order stays fixed, but the output order changes between the two families. This is the graph idea behind logarithmic inequalities; a later chapter turns it into a solving method.

Work it through

Worked example: compare inputs $1/4$ and 4. For base 2, $\log_2(1/4)=-2$ and $\log_2 4=2$ , so the larger input gives the larger output. For base $1/2$ , the values are 2 and $-2$ . Indeed, $\log_{1/2}x=-\log_2 x$ for all $x>0$ : the two graphs are reflections across the horizontal axis, with the same domain and range.

Pause. Think. Try.

1. If

0<u<v

, which statement is true?

\log_{1/3}u<\log_{1/3}v

\log_{1/3}u>\log_{1/3}v

\log_{1/3}u=\log_{1/3}v

Neither logarithm is defined because the base is below 1.

A hint, please

Follow the curve from the smaller input to the larger input.

Read the answer and explanation

$\log_{1/3}u>\log_{1/3}v$

The base

1/3

lies between 0 and 1, so the logarithm is strictly decreasing. The smaller input has the larger output.

2. As

x

approaches 0 through positive values, what happens to

\log_{1/2}x

?It approaches 0.It becomes undefined at every sufficiently small positive

x

.It decreases without bound.It increases without bound.

A hint, please

Use inputs

(1/2)^n

n

becomes a larger positive integer.

Read the answer and explanation

It increases without bound.

\log_{1/2}((1/2)^n)=n

. The input tends to 0 while the output grows without bound. The logarithm is still undefined at exactly 0.

About 5 minutes

TOPIC 03Class 11 · visual exploration

Swap the coordinates to see the inverse

The equations $y=b^x$ and $y=\log_b x$ reverse one another. If $(u,v)$ lies on the exponential graph, then $(v,u)$ lies on the logarithm graph. An input and its output exchange jobs.

Swapping coordinates reflects a graph in the line $y=x$ . To see that reflection faithfully, use the same physical scale on both axes: one unit horizontally must have the same length as one unit vertically. A stretched display can make a correct inverse pair look misleading.

The domains and ranges swap too. For $b^x$ , the input can be any real number and the output is positive. For $\log_b x$ , the input is positive and the output can be any real number.

The exponential’s horizontal asymptote $y=0$ becomes the logarithm’s vertical asymptote $x=0$ . Neither curve includes points on its own asymptote. In the explorer, select one point, read both coordinates and then identify its partner. The coordinate pair is the evidence; the picture helps you see the relationship.

Work it through

Worked example: pair points for base 2. Because $2^3=8$ , the exponential contains $(3,8)$ , while the logarithm contains $(8,3)$ . Also, $2^{-2}=1/4$ pairs $(-2,1/4)$ with $(1/4,-2)$ . Finally, $(0,1)$ on the exponential pairs with $(1,0)$ on the logarithm. In every pair, the coordinates swap; neither coordinate is negated.

Enable JavaScript to explore the inputs. The explanation and worked example remain available here.

Pause. Think. Try.

1. The point

(-3,1/8)

lies on

y=2^x

. Which point lies on its inverse?

(3,8)

(-1/8,3)

(1/8,-3)

(-3,8)

A hint, please

An inverse exchanges input and output.

Read the answer and explanation

$(1/8,-3)$

The inverse is

y=\log_2 x

. Swapping the coordinates gives

(1/8,-3)

, consistent with

\log_2(1/8)=-3

2. Why should the two axes use equal unit lengths when showing reflection in

y=x

?Otherwise the logarithm’s domain changes.Otherwise a correct inverse pair can look geometrically distorted.Because exponential graphs cannot use decimal inputs.Because the logarithm must have a positive output.

A hint, please

The coordinates remain correct, but a visual reflection depends on distance.

Read the answer and explanation

Otherwise a correct inverse pair can look geometrically distorted.

Unequal scales stretch the drawing differently in the two directions. The functions are still inverses, but equal unit lengths make their reflection in

y=x

visually faithful.

About 6 minutes

TOPIC 04Class 11 · functions and graphs

Move the curve; move its boundary

Now explore $g(x)=\log_2(x-2)+1$ . Begin with the argument, not the picture: $x-2>0$ , so $x>2$ . The vertical boundary is now $x=2$ .

Compare this with the parent $f(x)=\log_2 x$ . To feed the parent an input $u$ , the new graph needs $x-2=u$ , or $x=u+2$ . Then it adds 1 to the parent’s output. Every parent point $(u,v)$ becomes $(u+2,v+1)$ : two units right and one unit up.

This explains the direction of the horizontal shift. The minus sign inside the argument does not send the curve left. It means the input must be two larger to reach the same place in the logarithm.

The range remains all real numbers. Adding 1 to a set containing every real output still gives every real output. A horizontal shift changes the domain boundary; a vertical shift changes the heights. Use both the argument inequality and two transformed anchor points to check your sketch.

Work it through

Worked example: locate the intercept of $g(x)=\log_2(x-2)+1$ . Set the output to zero:

\log_2(x-2)=-1\quad\Longrightarrow\quad x-2=2^{-1}=\tfrac12\quad\Longrightarrow\quad x=\tfrac52.

It passes the domain check $x>2$ , so the intercept is $(5/2,0)$ . The parent points $(1,0)$ and $(2,1)$ also become $(3,1)$ and $(4,2)$ . There is no vertical-axis intercept because $x=0$ is outside the domain.

Pause. Think. Try.

1. For

y=\log_2(x-2)+1

, which pair is correct?Domain

x>-2

; asymptote

x=-2

.Domain

x>2

; asymptote

x=2

.Domain

x>2

; asymptote

y=1

.Domain

x\ge2

; asymptote

x=2

A hint, please

Solve the strict argument inequality.

Read the answer and explanation

Domain $x>2$ ; asymptote $x=2$ .

x-2>0

gives

x>2

. As the argument approaches 0 from the right, the vertical asymptote is

x=2

; the endpoint is excluded.

2. The parent graph

y=\log_2 x

contains

(1/2,-1)

. Where does that point move on

y=\log_2(x-2)+1

(-3/2,0)

(5/2,-2)

(1/2,0)

(5/2,0)

A hint, please

Add 2 to the input coordinate and 1 to the output coordinate.

Read the answer and explanation

$(5/2,0)$

(1/2+2,-1+1)=(5/2,0)

. Substitution gives

\log_2(1/2)+1=-1+1=0

About 6 minutes

TOPIC 05Class 11 · functions and graphs

Inside or outside: the minus sign matters

Compare $p(x)=-\log_2 x$ with $q(x)=\log_2(-x)$ . They look similar, but the minus signs act in different places.

In $p$ , the logarithm still receives $x>0$ . The output is then negated, so each point $(u,v)$ of the parent becomes $(u,-v)$ : a reflection across the horizontal axis. This is also the graph of $\log_{1/2}x$ .

In $q$ , the argument must satisfy $-x>0$ , which means $x<0$ . Each parent point $(u,v)$ becomes $(-u,v)$ : a reflection across the vertical axis. Negative inputs are allowed here because the whole argument $-x$ is positive.

Both graphs have range $\mathbb R$ and the vertical asymptote $x=0$ , but they lie on opposite sides of that boundary. Both are decreasing on their own domains. Do not join their pieces across zero or assume that a decreasing graph must lie below the horizontal axis.

For a more complicated expression, return to the same habit: inspect the argument, locate the zero of that argument and then map a few points.

Work it through

Worked example: three points on $q(x)=\log_2(-x)$ . At $x=-4$ , the argument is 4 and the output is 2. At $x=-1$ , the argument is 1 and the output is 0. At $x=-1/4$ , the argument is $1/4$ and the output is $-2$ . As the inputs increase from $-4$ to $-1/4$ , the outputs decrease from 2 to $-2$ . All three inputs satisfy $x<0$ .

Pause. Think. Try.

1. What is the domain of

y=\log_2(-x)

x>0

x<0

Every real

x

.No real input is allowed.

A hint, please

It is the argument

-x

, not necessarily

x

, that must be positive.

Read the answer and explanation

$x<0$

-x>0

is equivalent to

x<0

. For example, input

-2

gives the valid argument 2.

2. Which graph is the reflection of

y=\log_2 x

in the horizontal axis?

y=\log_2(-x)

y=\log_2(x-1)

y=-\log_2 x

y=2^x

A hint, please

A horizontal-axis reflection changes the sign of the output.

Read the answer and explanation

$y=-\log_2 x$

Negating the output gives

y=-\log_2 x

, with the same positive-input domain. Negating the input instead reflects in the vertical axis.

Write the reasoning

Show the domain, the key steps and a check of your answer.

Read a shifted and reflected graph

For $f(x)=2-\log_3(x+1)$ , find its domain, range, vertical asymptote, both intercepts and direction of change.

Compare your solution

The argument condition $x+1>0$ gives domain $(-1,\infty)$ , with vertical asymptote $x=-1$ . A logarithm has every real output, and subtracting it from 2 keeps the range $\mathbb R$ .

At $x=0$ , $f(0)=2-\log_3 1=2$ , giving vertical-axis intercept $(0,2)$ . For the horizontal-axis intercept, solve $\log_3(x+1)=2$ , so $x+1=9$ and $x=8$ . This is allowed, giving $(8,0)$ .

As $x$ increases, $x+1$ and its base-3 logarithm increase. Subtracting that logarithm makes $f$ strictly decreasing. Near $x=-1$ from the right, its output rises without bound; as $x$ grows without bound, its output falls without bound.

Build an inverse from a reflected logarithm

Let $g(x)=\log_2(4-x)$ . Find its domain and range, then find its inverse. Use the point $(0,2)$ to check the inverse pair.

Compare your solution

The argument requires $x<4$ , so the domain is $(-\infty,4)$ . The positive argument $4-x$ can take every positive value, giving range $\mathbb R$ .

Write $y=\log_2(4-x)$ . Converting to exponential form gives $2^y=4-x$ , hence $x=4-2^y$ . Swap the input and output names to obtain $g^{-1}(x)=4-2^x$ . Its domain is $\mathbb R$ and range is $(-\infty,4)$ , exactly the original range and domain exchanged.

Since $g(0)=\log_2 4=2$ , the inverse should contain $(2,0)$ . Indeed, $g^{-1}(2)=4-4=0$ . On equal-scale axes, the two points reflect in $y=x$ .

Class 11 milestone

You can read the curve before solving an equation

You can locate a logarithm’s permitted inputs, predict its direction and track shifts or reflections. Revisit the inverse-point explorer for a visual check, then continue to logarithmic equations and domain checks.

Follow the official syllabus

Logarithm Graphs: Follow the Input, Watch the Output

Start with three points and one boundary

Work it through

Pause. Think. Try.

A base below 1 turns the graph around

Work it through

Pause. Think. Try.

Swap the coordinates to see the inverse

Work it through

Pause. Think. Try.

Move the curve; move its boundary

Work it through

Pause. Think. Try.

Inside or outside: the minus sign matters

Work it through

Pause. Think. Try.

Your chapter checkpoint

Write the reasoning

Read a shifted and reflected graph

Build an inverse from a reflected logarithm

You can read the curve before solving an equation