THE WHOLE UNIT · ONE REFERENCE

Logarithms
Cheat sheet.

The key rules, formulas and reminders from all 15 topics, gathered into reference cards.

Key formulas, conditions and traps · Read down each column.

Defining the Exponential Function

Definition

$f(x) = a^x$ , with $a > 0$ and $a \ne 1$ .

The base is fixed; the variable is the power. That is the whole difference from $x^3$ .

Expression	Base	Power
$x^3$	varies	fixed
$3^x$	fixed	varies

Graphing Exponential Functions

Three points and one line

Point	Why
$\left(-1, \dfrac{1}{a}\right)$	a negative power is a reciprocal
$(0, 1)$	$a^0 = 1$ , for every base
$(1, a)$	$a^1 = a$

Asymptote: the $x$ -axis, $y = 0$ . The curve approaches it and never meets it.

Domain and range

Domain: every real number. Range: $y > 0$ .

Defining the Logarithm

Definition

$\log_a x = y \iff a^y = x$ , with $a > 0$ , $a \ne 1$ and $x > 0$ .

Out loud: $\log_a x$ is the power that turns $a$ into $x$ .

What is allowed

Part	Condition	Why
base $a$	$a > 0$ , $a \ne 1$	as for the exponential it inverts
argument $x$	$x > 0$	no power of a positive base is $0$ or negative
answer $y$	anything	a logarithm may be negative or fractional

The Inverse of an Exponential Function

The move

$x$ is stuck in an exponent, so the logarithm frees it.

y = a^x \iff x = \log_a y

Watch for

Adding a constant to $a^x$ moves the inverse's restriction: $f(x) = 2^x + 5$ has $f^{-1}$ defined only for $x > 5$ .
The graphs are mirror images in the line $y = x$ .

The Inverse of a Logarithmic Function

The move

$x$ is stuck inside a logarithm, so the exponential frees it.

y = \log_a x \iff x = a^y

Conditions on the Base and the Argument

The three conditions

\log_a x \quad\text{needs}\quad a > 0, \quad a \ne 1, \quad x > 0

The value has no restriction. It may be negative, fractional, or zero.

Traps

$x = 3$ is not allowed in $\log_5(x - 3)$ . The argument must be strictly positive, not zero.
A negative $x$ is fine if the argument still comes out positive, as in $\log_2(x^2 - 9)$ at $x = -4$ .
Not every logarithm restricts the domain. Check the argument before assuming it does.

The Four Basic Rules

The four

\log_a 1 = 0

\log_a a = 1

\log_a a^n = n

a^{\log_a x} = x

The third contains the first two: set $n = 0$ and $n = 1$ .

Traps

$\log_a 1 = 0$ , not $1$ . This is the most-swapped pair with $\log_a a = 1$ .
Different bases cannot be combined. Evaluate each term, then add.
$3^{\log_3 4 + 1} = 3^{\log_3 4} \cdot 3 = 4 \cdot 3 = 12$ . Split the exponent first.

When the Base Is Not Written

The move

No base written means base $10$ .

\log x = \log_{10} x

Watch out

Only $\log$ hides a base. $\log_2 8$ keeps its $2$ .
$x$ must still be positive: $\log 0$ and $\log(-5)$ have no value.
A number under $1$ gives a negative answer, never an undefined one.

The Natural Logarithm

The move

$\ln$ is a logarithm in base $e$ , where $e \approx 2.71828$ .

\ln x = \log_e x

Watch out

$\ln$ has no $e$ written, but it is there. $\ln e^3 = 3$ , not $3e$ .
Domain unchanged: $\ln 0$ and $\ln(-4)$ have no value.
$\ln x$ is negative for $0 < x < 1$ , not undefined.
$e$ never has to be turned into a decimal to answer these.

The Product Rule

The rule

\log_a x + \log_a y = \log_a(xy)

Same base only. Adding logarithms multiplies their arguments.

Traps

\log_a x + \log_a y \ne \log_a(x + y)

\log_a x \cdot \log_a y \ne \log_a(xy)

The Quotient Rule

The rule

\log_a x - \log_a y = \log_a \frac{x}{y}

Same base only. The first argument goes on top.

Traps

\log_a x - \log_a y \ne \log_a(x - y)

\frac{\log_a x}{\log_a y} \ne \log_a \frac{x}{y}

Order matters: reversing the two terms flips the sign of the answer.

Swapping the Base and the Exponent

The three

\log_a x^m = m \log_a x

\log_{a^n} x = \frac{1}{n} \log_a x

\log_{a^n} x^m = \frac{m}{n} \log_a x

Argument exponent multiplies. Base exponent divides.

Traps

$\log_a x^m$ is not $(\log_a x)^m$ . Compare $\log_2 4^3 = 6$ with $(\log_2 4)^3 = 8$ .
The base exponent goes underneath: $\log_{a^n} x = \dfrac{1}{n} \log_a x$ , never $n \log_a x$ .
Roots are fractional powers: $\log_4 \sqrt[3]{16} = \dfrac{4/3}{2} = \dfrac{2}{3}$ .
The strict form is $\log_a x^2 = 2 \log_a |x|$ , because $x$ may be negative.

When a Logarithm Is the Exponent

The three

a^{\log_a x} = x

a^{m \log_a x} = x^m

a^{\log_b c} = c^{\log_b a}

The first needs matching bases. The second is the first with an outside power. The third trades the two ends when nothing matches.

Traps

$2^{\log_3 7}$ cancels to nothing. The bases share no common power here.
A negative exponent is a reciprocal, not a negative answer: $3^{-2 \log_3 2} = 2^{-2} = \dfrac{1}{4}$ .
$2^{\log_2 5 + 3}$ is $40$ , not $8$ . The $3$ multiplies, it does not vanish.
$a^{m \log_a x} = x^m$ , never $m x$ .

Changing the Base

The rule

\log_a x = \frac{\log_b x}{\log_b a}

Argument on top, old base underneath. The new base $b$ is yours to choose.

Traps

$\dfrac{\ln 3}{\ln 7}$ is $\log_7 3$ , not $\log_3 7$ . Check which one is underneath.
$\dfrac{\log_a x}{\log_a y}$ is not $\log_a \dfrac{x}{y}$ . That is the quotient rule, a different thing.

Graphing Logarithmic Functions

The shape

$y = \log_a x$ : domain $x > 0$ , range every real $y$ , vertical asymptote $x = 0$ .

Base	Shape
$a > 1$	rises, slowly and forever
$0 < a < 1$	falls, slowly and forever

The two are mirror images in the $x$ -axis, because $\log_{1/a} x = -\log_a x$ .

Traps

A negative height never means a negative $x$ . $\log_3 x = -2$ gives $x = \dfrac{1}{9}$ , not $-9$ .
Heights are negative exactly on $0 < x < 1$ .
The curve is unbounded below near the wall, but never crosses it.
$\log_2 x$ gains one unit of height each time $x$ doubles.

Defining the Exponential Function

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Definition

$f(x) = a^x$ , with $a > 0$ and $a \ne 1$ .

The base is fixed; the variable is the power. That is the whole difference from $x^3$ .

Expression	Base	Power
$x^3$	varies	fixed
$3^x$	fixed	varies

Why the two conditions

Condition	What it prevents
$a > 0$	$a^{1/2} = \sqrt{a}$ has no real value for a negative base
$a \ne 1$	$1^x = 1$ is a flat line, with no inverse

The rules of indices, unchanged

$a^x \cdot a^y = a^{x + y}$ , $\dfrac{a^x}{a^y} = a^{x - y}$ , $\left(a^x\right)^y = a^{xy}$

$a^0 = 1$ , and $a^{-x} = \dfrac{1}{a^x}$ - a negative power is a reciprocal, never a negative value.

What the graph does

Fact	Consequence
$a^x > 0$ always	the curve never touches the $x$ -axis
$a^0 = 1$	every exponential passes through $(0, 1)$
$a > 1$	growth: climbs without limit
$0 < a < 1$	decay: falls towards $0$

Solving $a^{\text{something}} = a^{\text{something}}$

Put both sides on the same base, then equate the powers. An exponential takes each value once, so nothing is lost.

Graphing Exponential Functions

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Three points and one line

Point	Why
$\left(-1, \dfrac{1}{a}\right)$	a negative power is a reciprocal
$(0, 1)$	$a^0 = 1$ , for every base
$(1, a)$	$a^1 = a$

Asymptote: the $x$ -axis, $y = 0$ . The curve approaches it and never meets it.

Which shape

Base	Shape
$a > 1$	growth: rises left to right
$0 < a < 1$	decay: falls left to right

Domain and range

Domain: every real number. Range: $y > 0$ .

Two things that move it

Change	Effect
$\left(\dfrac{1}{a}\right)^x = a^{-x}$	reflection in the $y$ -axis - decay is growth mirrored
$a^x + k$	every point lifts by $k$ , and the asymptote goes to $y = k$

Defining the Logarithm

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Definition

$\log_a x = y \iff a^y = x$ , with $a > 0$ , $a \ne 1$ and $x > 0$ .

Out loud: $\log_a x$ is the power that turns $a$ into $x$ .

Swapping the two forms

Index form	Logarithm form
$3^4 = 81$	$\log_3 81 = 4$
$2^{-3} = \dfrac{1}{8}$	$\log_2 \dfrac{1}{8} = -3$
$9^{1/2} = 3$	$\log_9 3 = \dfrac{1}{2}$

The base stays the base. Only the other two change places.

Free values

$\log_a 1 = 0$ and $\log_a a = 1$ , for every allowed base.

What is allowed

Part	Condition	Why
base $a$	$a > 0$ , $a \ne 1$	as for the exponential it inverts
argument $x$	$x > 0$	no power of a positive base is $0$ or negative
answer $y$	anything	a logarithm may be negative or fractional

Undoing

$\log_a(a^x) = x$ and $a^{\log_a x} = x$ . The graphs are mirror images in $y = x$ .

The Inverse of an Exponential Function

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The move

$x$ is stuck in an exponent, so the logarithm frees it.

y = a^x \iff x = \log_a y

Method

Write $y = f(x)$ .
Peel off everything outside the power first.
Take $\log_a$ of both sides. It releases the whole exponent.
Undo what is left, then swap the letters.

Function	Inverse
$f(x) = 2^x$	$f^{-1}(x) = \log_2 x$
$f(x) = 2^x + 5$	$f^{-1}(x) = \log_2(x - 5)$
$f(x) = 3 \cdot 2^x$	$f^{-1}(x) = \log_2 \dfrac{x}{3}$
$f(x) = 2^{x - 1}$	$f^{-1}(x) = \log_2 x + 1$
$f(x) = 5^{2x}$	$f^{-1}(x) = \dfrac{\log_5 x}{2}$

The whole exponent comes out

$\log_2\!\left(2^{3x + 1}\right) = 3x + 1$ , not $3x$ . Take the exponent out in one piece, then unpick it.

Domain and range change places

	domain	range
$f(x) = a^x$	every real number	$y > 0$
$f^{-1}(x) = \log_a x$	$x > 0$	every real number

Watch for

Adding a constant to $a^x$ moves the inverse's restriction: $f(x) = 2^x + 5$ has $f^{-1}$ defined only for $x > 5$ .
The graphs are mirror images in the line $y = x$ .

The Inverse of a Logarithmic Function

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The move

$x$ is stuck inside a logarithm, so the exponential frees it.

y = \log_a x \iff x = a^y

Method

Write $y = f(x)$ .
Make the logarithm stand alone: clear coefficients and constants first.
Raise the base to both sides. It releases the whole argument.
Undo what is left, then swap the letters.

Function	Inverse
$f(x) = \log_3 x$	$f^{-1}(x) = 3^x$
$f(x) = \log_2 x + 1$	$f^{-1}(x) = 2^{x - 1}$
$f(x) = \log_5(x - 2)$	$f^{-1}(x) = 5^x + 2$
$f(x) = 2\log_3 x$	$f^{-1}(x) = 3^{x/2}$
$f(x) = \log_4(3x)$	$f^{-1}(x) = \dfrac{4^x}{3}$

Stand alone first

$2^{3\log_2 x} \ne x$ . Divide by the $3$ before raising the base:

3\log_2 x = y \;\Rightarrow\; \log_2 x = \frac{y}{3} \;\Rightarrow\; x = 2^{y/3}

The whole argument comes out

$5^{\log_5(x - 2)} = x - 2$ . The bracket arrives intact; the $-2$ comes off afterwards.

Domain and range change places

	domain	range
$f(x) = \log_a x$	$x > 0$	every real number
$f^{-1}(x) = a^x$	every real number	$y > 0$

Which way round

$x$ sits...	...so use
in an exponent	the logarithm
inside a logarithm	the exponential

Conditions on the Base and the Argument

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The three conditions

\log_a x \quad\text{needs}\quad a > 0, \quad a \ne 1, \quad x > 0

The value has no restriction. It may be negative, fractional, or zero.

Where each condition comes from

Condition	Reason
$a > 0$	a negative base breaks between whole powers: $(-4)^{1/2}$ is not real
$a \ne 1$	every power of $1$ is $1$ , so $1^y = x$ has no single answer
$x > 0$	a positive base to any real power is positive, never $0$ or negative

Reading conditions off an expression

One inequality per condition, then keep what they all allow.

Expression	Conditions	Allowed
$\log_3(x - 2)$	$x - 2 > 0$	$x > 2$
$\log_5(7 - x)$	$7 - x > 0$	$x < 7$
$\log_2(x^2 - 9)$	$x^2 - 9 > 0$	$x < -3$ or $x > 3$
$\log_4(x^2 + 1)$	always true	every real $x$
$\log_x 9$	$x > 0$ , $x \ne 1$	$x > 0$ , $x \ne 1$
$\log_{x-4} 5$	$x - 4 > 0$ , $x - 4 \ne 1$	$x > 4$ , $x \ne 5$

When the base contains x

Both base conditions apply, and the argument condition applies as well. Three inequalities.

\log_{x-1}(6 - 2x): \quad x > 1, \quad x \ne 2, \quad x < 3 \;\Rightarrow\; 1 < x < 3, ; x \ne 2

Traps

$x = 3$ is not allowed in $\log_5(x - 3)$ . The argument must be strictly positive, not zero.
A negative $x$ is fine if the argument still comes out positive, as in $\log_2(x^2 - 9)$ at $x = -4$ .
Not every logarithm restricts the domain. Check the argument before assuming it does.

The Four Basic Rules

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The four

\log_a 1 = 0

\log_a a = 1

\log_a a^n = n

a^{\log_a x} = x

The third contains the first two: set $n = 0$ and $n = 1$ .

The method

Write the argument as a power of the base, then read the exponent off.

Logarithm	As a power	Value
$\log_2 32$	$2^5$	$5$
$\log_2 64$	$2^6$	$6$
$\log_3 \dfrac{1}{81}$	$3^{-4}$	$-4$
$\log_2 \sqrt{2}$	$2^{1/2}$	$\dfrac{1}{2}$
$\log_5 \dfrac{1}{\sqrt{5}}$	$5^{-1/2}$	$-\dfrac{1}{2}$
$\log_4 1$	$4^0$	$0$
$\log_7 7$	$7^1$	$1$

Roots and fractions are not special cases:

\sqrt[n]{a} = a^{1/n},

\frac{1}{a^n} = a^{-n}

Cancelling

$a^{\log_a x} = x$ only when the bases match.

3^{\log_3 7} = 7

\text{but}

2^{\log_3 7} \text{ simplifies to nothing}

When the argument is not a whole-number power

Go back to index form and put both sides on a common base.

\log_4 8 = y \;\Rightarrow\; 4^y = 8 \;\Rightarrow\; 2^{2y} = 2^3 \;\Rightarrow\; y = \frac{3}{2}

Traps

$\log_a 1 = 0$ , not $1$ . This is the most-swapped pair with $\log_a a = 1$ .
Different bases cannot be combined. Evaluate each term, then add.
$3^{\log_3 4 + 1} = 3^{\log_3 4} \cdot 3 = 4 \cdot 3 = 12$ . Split the exponent first.

When the Base Is Not Written

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The move

No base written means base $10$ .

\log x = \log_{10} x

Read the exponent off

$x$	$10^n$	$\log x$
$10000$	$10^4$	$4$
$1000$	$10^3$	$3$
$100$	$10^2$	$2$
$10$	$10^1$	$1$
$1$	$10^0$	$0$
$0.1$	$10^{-1}$	$-1$
$0.01$	$10^{-2}$	$-2$
$0.001$	$10^{-3}$	$-3$

The four rules with the ten hidden

Rule	Example
$\log 1 = 0$	$\log 1 = 0$
$\log 10 = 1$	$\log 10 = 1$
$\log 10^n = n$	$\log 10^6 = 6$
$10^{\log x} = x$	$10^{\log 7} = 7$

Sanity check any answer

Trap the number between the powers of ten either side.

Because	$\log x$ lies between
$10 < 50 < 100$	$1$ and $2$
$100 < 700 < 1000$	$2$ and $3$
$1 < 5 < 10$	$0$ and $1$
$0.01 < 0.04 < 0.1$	$-2$ and $-1$

Watch out

Only $\log$ hides a base. $\log_2 8$ keeps its $2$ .
$x$ must still be positive: $\log 0$ and $\log(-5)$ have no value.
A number under $1$ gives a negative answer, never an undefined one.

The Natural Logarithm

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The move

$\ln$ is a logarithm in base $e$ , where $e \approx 2.71828$ .

\ln x = \log_e x

Three notations

Written	Base
$\log_a x$	$a$ , stated
$\log x$	$10$ , hidden
$\ln x$	$e$ , hidden

The four rules with a = e

Rule	Why
$\ln 1 = 0$	$e^0 = 1$
$\ln e = 1$	$e^1 = e$
$\ln e^n = n$	the exponent is the answer
$e^{\ln x} = x$	the two cancel

Read at sight

$x$	$\ln x$
$e^7$	$7$
$e^2$	$2$
$e$	$1$
$1$	$0$
$\sqrt{e}$	$\dfrac{1}{2}$
$e^{1/3}$	$\dfrac{1}{3}$
$\dfrac{1}{e}$	$-1$
$\dfrac{1}{e^5}$	$-5$

Method

Write the argument as a power of $e$ .
Read off the exponent.
A root is a fractional power; a reciprocal is a negative one.

Watch out

$\ln$ has no $e$ written, but it is there. $\ln e^3 = 3$ , not $3e$ .
Domain unchanged: $\ln 0$ and $\ln(-4)$ have no value.
$\ln x$ is negative for $0 < x < 1$ , not undefined.
$e$ never has to be turned into a decimal to answer these.

The Product Rule

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The rule

\log_a x + \log_a y = \log_a(xy)

Same base only. Adding logarithms multiplies their arguments.

Why

A logarithm is an exponent, and exponents add when powers multiply.

a^m \cdot a^n = a^{m+n}

Worth knowing by sight

Sum	Single logarithm	Value
$\log_2 4 + \log_2 8$	$\log_2 32$	$5$
$\log 25 + \log 4$	$\log 100$	$2$
$\log_3 6 + \log_3 4.5$	$\log_3 27$	$3$
$\log_6 4 + \log_6 9$	$\log_6 36$	$2$
$\log_2 12 + \log_2 \dfrac{4}{3}$	$\log_2 16$	$4$

Neither term has to be a whole number. Only the product does.

Splitting, the other direction

Break the argument into factors so one of them is a power of the base.

\log_2 40 = \log_2(8 \cdot 5) = 3 + \log_2 5

The bases must match

Expression	Combines?
$\log_5 3 + \log_5 7$	yes, $\log_5 21$
$\log 4 + \log 25$	yes, both base $10$
$\ln 2 + \ln 6$	yes, both base $e$
$\log_5 3 + \log_2 7$	no

Traps

\log_a x + \log_a y \ne \log_a(x + y)

\log_a x \cdot \log_a y \ne \log_a(xy)

Given-value questions

Build the argument from the numbers you were given, then convert the product into a sum.

\log_a 2 = 0.3,\; \log_a 5 = 0.7 \;\Rightarrow\; \log_a 10 = 0.3 + 0.7 = 1

The Quotient Rule

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The rule

\log_a x - \log_a y = \log_a \frac{x}{y}

Same base only. The first argument goes on top.

Why

Exponents subtract when powers divide.

\frac{a^m}{a^n} = a^{m-n}

The reciprocal shortcut

Put $x = 1$ and use $\log_a 1 = 0$ :

\log_a \frac{1}{y} = -\log_a y

So $\log_2 \dfrac{1}{8} = -3$ , $\log_3 \dfrac{1}{9} = -2$ , $\ln \dfrac{1}{e} = -1$ . This is why the whole curve between $0$ and $1$ sits below the axis.

Worth knowing by sight

Expression	Single logarithm	Value
$\log_2 48 - \log_2 3$	$\log_2 16$	$4$
$\log 5000 - \log 5$	$\log 1000$	$3$
$\ln e^7 - \ln e^4$	$\ln e^3$	$3$
$\log_2 5 - \log_2 40$	$\log_2 \dfrac{1}{8}$	$-3$
$\log_2 20 - \log_2 5$	$\log_2 4$	$2$

Both rules together

Sums go on top, differences go underneath.

\log_a 6 + \log_a 5 - \log_a 3 = \log_a \frac{6 \cdot 5}{3} = \log_a 10

\log_6 4 + \log_6 27 - \log_6 18 = \log_6 \frac{108}{18} = \log_6 6 = 1

Traps

\log_a x - \log_a y \ne \log_a(x - y)

\frac{\log_a x}{\log_a y} \ne \log_a \frac{x}{y}

Order matters: reversing the two terms flips the sign of the answer.

Given-value questions

\log_a 3 = 0.5,\; \log_a 2 = 0.3 \;\Rightarrow\; \log_a 1.5 = \log_a \frac{3}{2} = 0.5 - 0.3 = 0.2

Swapping the Base and the Exponent

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The three

\log_a x^m = m \log_a x

\log_{a^n} x = \frac{1}{n} \log_a x

\log_{a^n} x^m = \frac{m}{n} \log_a x

Argument exponent multiplies. Base exponent divides.

Where each comes from

Rule	Derivation
$\log_a x^m = m \log_a x$	$x = a^k \Rightarrow x^m = a^{km}$
$\log_{a^n} x = \dfrac{1}{n} \log_a x$	$(a^n)^y = a^{ny} = x \Rightarrow ny = \log_a x$

The method

Write the base and the argument as powers of one common number.
Divide the argument exponent by the base exponent.
Check by raising the base back up.

Expression	As powers	Value
$\log_2 8^5$	$2^{15}$ in the argument	$15$
$\log_3 9^4$	$3^8$ in the argument	$8$
$\log_4 8$	$\log_{2^2} 2^3$	$\dfrac{3}{2}$
$\log_9 27$	$\log_{3^2} 3^3$	$\dfrac{3}{2}$
$\log_8 4$	$\log_{2^3} 2^2$	$\dfrac{2}{3}$
$\log_8 32$	$\log_{2^3} 2^5$	$\dfrac{5}{3}$
$\log_{27} 81$	$\log_{3^3} 3^4$	$\dfrac{4}{3}$
$\log_{16} 8$	$\log_{2^4} 2^3$	$\dfrac{3}{4}$
$\log_9 \sqrt{3}$	$\log_{3^2} 3^{1/2}$	$\dfrac{1}{4}$

Bases below 1

A fractional base is a negative power, so the value turns negative.

Expression	As powers	Value
$\log_{1/2} 8$	$\log_{2^{-1}} 2^3$	$-3$
$\log_{1/3} 9$	$\log_{3^{-1}} 3^2$	$-2$
$\log_{1/4} 8$	$\log_{2^{-2}} 2^3$	$-\dfrac{3}{2}$

Traps

$\log_a x^m$ is not $(\log_a x)^m$ . Compare $\log_2 4^3 = 6$ with $(\log_2 4)^3 = 8$ .
The base exponent goes underneath: $\log_{a^n} x = \dfrac{1}{n} \log_a x$ , never $n \log_a x$ .
Roots are fractional powers: $\log_4 \sqrt[3]{16} = \dfrac{4/3}{2} = \dfrac{2}{3}$ .
The strict form is $\log_a x^2 = 2 \log_a |x|$ , because $x$ may be negative.

When a Logarithm Is the Exponent

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The three

a^{\log_a x} = x

a^{m \log_a x} = x^m

a^{\log_b c} = c^{\log_b a}

The first needs matching bases. The second is the first with an outside power. The third trades the two ends when nothing matches.

The method

Write the base of the power as a power of the base of the logarithm.
Bring the new multiplier out as an outside power.
Cancel the matching pair and raise what is left.

8^{\log_2 5} = 2^{3 \log_2 5} = \left(2^{\log_2 5}\right)^3 = 5^3 = 125

Read at sight

Expression	Rewritten	Value
$5^{\log_5 9}$	already matched	$9$
$3^{2 \log_3 4}$	$4^2$	$16$
$e^{2 \ln 3}$	$3^2$	$9$
$4^{\log_2 3}$	$2^{2 \log_2 3}$	$9$
$9^{\log_3 5}$	$3^{2 \log_3 5}$	$25$
$25^{\log_5 3}$	$5^{2 \log_5 3}$	$9$
$8^{\log_2 5}$	$2^{3 \log_2 5}$	$125$
$\left(\dfrac{1}{9}\right)^{\log_3 2}$	$3^{-2 \log_3 2}$	$\dfrac{1}{4}$
$2^{\log_4 9}$	$2^{\frac{1}{2} \log_2 9}$	$3$

Extra terms in the exponent

Split the sum first, then cancel.

2^{\log_2 5 + 3} = 2^{\log_2 5} \cdot 2^3 = 40

5^{\log_5 2 - 1} = \frac{2}{5}

Swapping the ends

a^{\log_b c} = c^{\log_b a}

\text{so}

81^{\log_3 2} = 2^{\log_3 81} = 2^4 = 16

Traps

$2^{\log_3 7}$ cancels to nothing. The bases share no common power here.
A negative exponent is a reciprocal, not a negative answer: $3^{-2 \log_3 2} = 2^{-2} = \dfrac{1}{4}$ .
$2^{\log_2 5 + 3}$ is $40$ , not $8$ . The $3$ multiplies, it does not vanish.
$a^{m \log_a x} = x^m$ , never $m x$ .

Changing the Base

6 reference blocks

Read lesson ↗

The rule

\log_a x = \frac{\log_b x}{\log_b a}

Argument on top, old base underneath. The new base $b$ is yours to choose.

Why

$y = \log_a x$ means $a^y = x$ . Take $\log_b$ of both sides:

y \log_b a = \log_b x \;\Rightarrow\; y = \frac{\log_b x}{\log_b a}

Choosing the new base

Goal	Choose
use a calculator	$b = 10$ or $b = e$
get an exact value	a base both numbers are powers of

\log_2 7 = \frac{\ln 7}{\ln 2},

\log_8 32 = \frac{\log_2 32}{\log_2 8} = \frac{5}{3},

\log_9 27 = \frac{3}{2}

The upside-down case

Put $x = b$ and use $\log_b b = 1$ :

\log_a b = \frac{1}{\log_b a}

\Longrightarrow

\log_a b \cdot \log_b a = 1

So $\log_5 3 \cdot \log_3 5 = 1$ , and if $\log_a b = 4$ then $\log_b a = \dfrac{1}{4}$ . Reciprocal, never negative.

Chains collapse

Put every factor over one base and the middles cancel.

\log_2 5 \cdot \log_5 8 = \frac{\ln 5}{\ln 2} \cdot \frac{\ln 8}{\ln 5} = \log_2 8 = 3

\log_2 3 \cdot \log_3 4 \cdot \log_4 5 \cdot \log_5 8 = \log_2 8 = 3

A chain keeps only the first base and the last argument.

Traps

$\dfrac{\ln 3}{\ln 7}$ is $\log_7 3$ , not $\log_3 7$ . Check which one is underneath.
$\dfrac{\log_a x}{\log_a y}$ is not $\log_a \dfrac{x}{y}$ . That is the quotient rule, a different thing.

Graphing Logarithmic Functions

6 reference blocks

Read lesson ↗

The shape

$y = \log_a x$ : domain $x > 0$ , range every real $y$ , vertical asymptote $x = 0$ .

Base	Shape
$a > 1$	rises, slowly and forever
$0 < a < 1$	falls, slowly and forever

The two are mirror images in the $x$ -axis, because $\log_{1/a} x = -\log_a x$ .

Three points fix the curve

\left(\frac{1}{a}, -1\right),

(1, 0),

(a, 1)

For $y = \log_2 x$ : $\left(\dfrac{1}{2}, -1\right)$ , $(1, 0)$ , $(2, 1)$ . Plot them, follow the shape, done.

$(1, 0)$ is on every logarithm graph, whatever the base.

Reflection in y = x

	$y = a^x$	$y = \log_a x$
domain	every real $x$	$x > 0$
range	$y > 0$	every real $y$
asymptote	horizontal, $y = 0$	vertical, $x = 0$
passes through	$(0, 1)$	$(1, 0)$

Every row is the row above with the coordinates swapped.

Shifts

y = \log_a(x - h) + k

asymptote $x = h$ , domain $x > h$ — the inside constant moves the wall
$+k$ lifts the curve and leaves the wall alone

Method for a shifted graph

Take $f(x) = \log_2(x - 3) + 1$ :

Wall where the argument is zero: $x = 3$ , so domain $x > 3$
Argument $= 1$ at $x = 4$ : $f(4) = 0 + 1 = 1$
Argument $= 2$ at $x = 5$ : $f(5) = 1 + 1 = 2$
Base above $1$ , so it rises from the wall through $(4, 1)$ and $(5, 2)$

Traps

A negative height never means a negative $x$ . $\log_3 x = -2$ gives $x = \dfrac{1}{9}$ , not $-9$ .
Heights are negative exactly on $0 < x < 1$ .
The curve is unbounded below near the wall, but never crosses it.
$\log_2 x$ gains one unit of height each time $x$ doubles.