Using Properties of Logarithms

Rule

Properties of Logarithms

The Properties of Logarithms allow expressions with logarithms to be rewritten.

Rule

Product Property of Logarithms

Rule

lo g_{b} (m n) = lo g_{b} (m) + lo g_{b} (n)

The logarithm of a product can be written as the sum of the logarithms of each factor in the product. For example,

lo g (7 \cdot 4)

can be expressed using this rule.

lo g (7 \cdot 4) = lo g (7) + lo g (4)

This rule can be explained using the identity

x = b^{l o g_{b} (x)} .

lo g_{b} (m n)

RewriteRewrite

m n

as

m \cdot n

lo g_{b} (m \cdot n)

m = b^{l o g_{b} (m)}

lo g_{b} (b^{l o g_{b} (m)} \cdot b^{l o g_{b} (n)})

MultPow

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

lo g_{b} (b^{l o g_{b} (m) + l o g_{b} (n)})

lo g_{b} (b^{m}) = m

lo g_{b} (m) + lo g_{b} (n)

This rule is valid for positive values of

b,

m,

and

n

and

b \neq = 1 .

Rule

Quotient Property of Logarithms

Rule

lo g_{b} (\frac{m}{n}) = lo g_{b} (m) - lo g_{b} (n)

The logarithm of a quotient can be written as the difference between the logarithm of the numerator and the logarithm of the denominator. For example,

lo g (\frac{7}{4})

can be expressed using this rule.

lo g (\frac{7}{4}) = lo g (7) - lo g (4)

This rule can be explained using the identity

x = b^{l o g_{b} (x)} .

lo g_{b} (\frac{m}{n})

m = b^{l o g_{b} (m)}

lo g_{b} (\frac{b ^{l o g_{b} (m)}}{b ^{l o g_{b} (n)}})

DivPow

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

lo g_{b} (b^{l o g_{b} (m) - l o g_{b} (n)})

lo g_{b} (b^{m}) = m

lo g_{b} (m) - lo g_{b} (n)

This rule is valid for positive values of

b,

m,

and

n

and

b \neq = 1 .

Rule

Power Property of Logarithms

Rule

lo g_{b} (m^{n}) = n lo g_{b} (m)

The logarithm of a power can be written as the product of the exponent and the logarithm of the base. For example,

lo g (7^{4})

can be expressed using this rule.

lo g (7^{4}) = 4 \cdot lo g (7)

This rule can be explained using the identity

x = b^{l o g_{b} (x)} .

lo g_{b} (m^{n})

m = b^{l o g_{b} (m)}

lo g_{b} ((b^{l o g_{b} (m)})^{n})

PowPow

(a^{m})^{n} = a^{m \cdot n}

lo g_{b} (b^{l o g_{b} (m) \cdot n})

lo g_{b} (b^{m}) = m

lo g_{b} (m) \cdot n

CommutativePropMultCommutative Property of Multiplication

n lo g_{b} (m)

This rule is valid for positive values of

b,

m,

and

n

and

b \neq = 1 .

Simplify the expression

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Exercise

Simplify the expression $lo g_{6} (8) - lo g_{6} (3) + lo g_{6} (9) - lo g_{6} (4)$ using the properties of logarithms.

Show Solution

Solution

The Quotient Property of Logarithms states that the following is true for logarithms.

lo g_{b} (\frac{m}{n}) = lo g_{b} (m) - lo g_{b} (n)

In our expression the first two terms and the last two terms both are a difference of two logarithms with the same base. We can use the Quotient Property of Logarithms to rewrite each pair into a logarithm of a quotient.

lo g_{6} (8) - lo g_{6} (3) + lo g_{6} (9) - lo g_{6} (4) ⇕ lo g_{6} (\frac{8}{3}) + lo g_{6} (\frac{9}{4})

This expression is a sum of two logarithms with the same base. We can write them into a logarithm of a product using the Product Property of Logarithms.

lo g_{6} (\frac{8}{3}) + lo g_{6} (\frac{9}{4}) \Leftrightarrow lo g_{6} (\frac{8}{3} \cdot \frac{9}{4})

Let's continue simplifying this expression.

lo g_{6} (\frac{8}{3} \cdot \frac{9}{4})

MultFracMultiply fractions

lo g_{6} (\frac{8 \cdot 9}{3 \cdot 4})

SimpQuotSimplify quotient

lo g_{6} (6)

lo g_{6} (6) = 1

1

Thus, the expression

lo g_{6} (8) - lo g_{6} (3) + lo g_{6} (9) - lo g_{6} (4)

equals

1 .

Rule

Change of Base Formula

The Change of Base Formula allows the logarithm of an arbitrary base to be rewritten as the quotient of two logarithms with another base. $lo g_{c} (a) = \frac{lo g _{b} ( a )}{lo g _{b} ( c )}$ With many calculators it is only possible to evaluate the common and the natural logarithm. The Change of Base Formula can then be used to evaluate logarithms of other bases.

lo g_{c} (a) = \frac{lo g ( a )}{lo g ( c )} and lo g_{c} (a) = \frac{ln ( a )}{ln ( c )}

Method

Solving an Exponential Equation using Logarithms

An exponential equation can be solved by applying logarithms and using the Power Property of Logarithms. Consider the following equation.

8^{x} = 3

1

Apply the logarithm

First, the equation is rewritten by applying a logarithm on both sides. $8^{x} = 3 \Leftrightarrow lo g (8^{x}) = lo g (3)$

2

Rewrite using the Power Property of Logarithms

By using the Power Property of Logarithms, powers can be rewritten into a product. $lo g (8^{x}) = lo g (3) ⇕ x \cdot lo g (8) = lo g (3)$

3

Solve the resulting equation

After the power has been rewritten into a product, the unknown variable can be isolated using inverse operations. Here, $x$ gets isolated on the left-hand side when both sides of the equation are divided by $lo g (8) .$ $x \cdot lo g (8) = lo g (3) ⇕ x = \frac{lo g ( 3 )}{lo g ( 8 )}$ By using a calculator, an approximate value of $x$ can be calculated. Here, $x \approx 0.53 .$

Solve the equation using the common logarithm

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Exercise

Solve the equation $4^{x} = 25$ using the common logarithm. State the answer with three significant digits.

Show Solution

Solution

The given equation can be solved by applying the

lo g_{4}

on both sides. Since many calculators are limited to only the common and the natural logarithms, it is often not possible to use

lo g_{4} .

We can solve it anyway. Let's begin by applying the common logarithm on both sides.

4^{x} = 25 \Leftrightarrow lo g (4^{x}) = lo g (25)

The Power Property of Logarithms gives us the relationship

lo g_{b} (m^{n}) = n \cdot lo g_{b} (m) .

With this we can rewrite the left-hand side of the equation.

lo g (4^{x}) = lo g (25) \Leftrightarrow x \cdot lo g (4) = lo g (25)

We can now solve the equation for

x

by dividing both sides of the equation with

lo g (4) .

x \cdot lo g (4) = lo g (25)

DivEqn

LHS / lo g (4) = RHS / lo g (4)

x = \frac{lo g ( 2 5 )}{lo g ( 4 )}

UseCalcUse a calculator

x = 2.32192 \dots

RoundDecRound to

x \approx 2.32

The solution to the equation is

x \approx 2.32 .

Rule

Inverse Properties of Logarithms

A logarithmic function $g (x) = lo g_{b} (x),$ is by definition the inverse of an exponential function $f (x) = b^{x} .$ This means that their function composition results in the identity function.

$g (f (x)) = lo g_{b} (b^{x}) = x$
$f (g (x)) = b^{l o g_{b} (x)} = x$

The fact that a logarithm and a power with the same base undo each other is what is known as the inverse properties of logarithms.

$lo g_{b} (b^{x}) = x and b^{l o g_{b} (x)} = x$

They also hold true for the common logarithm and the natural logarithm.

$lo g (1 0^{x}) = x and ln (e^{x}) = x and 1 0^{l o g (x)} = x e^{l n (x)} = x$

These properties together with other properties of logarithms permit to simplify logarithmic expressions and to solve equations involving logarithms and powers. Some particular examples are shown below.

$l n (e^{4 x}) 4 x x = 20 = 20 = 5 3^{l o g_{3} (5 x)} 5 x x = 10 = 10 = 2$

Solve the equation using the inverse properties of logarithms

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Exercise

Solve the equation $\frac{e ^{l n (3^{4})} \cdot 1 0 ^{l o g (x)}}{lo g _{5} ( 1 2 5 )} = lo g (100)$ using the inverse properties of logarithms.

Show Solution

Solution

The inverse properties of logarithms tell us that a logarithm and a power with the same base undo each other.

lo g_{b} (b^{x}) = x and b^{l o g_{b} (x)} = x

On the left-hand side, we can identify

ln

as being the logarithm with base

e

and

lo g

as the logarithm with base

10 .

Thus, we can simplify the factors

e^{l n (3^{4})}

and

1 0^{l o g (x)}

using the inverse properties of logarithms.

\frac{e ^{l n (3^{4})} \cdot 1 0 ^{l o g (x)}}{lo g _{5} ( 1 2 5 )} ⇕ \frac{3 ^{4} \cdot x}{lo g _{5} ( 1 2 5 )} = lo g (100) = lo g (100)

To simplify the remaining two logarithms, we first need to rewrite their arguments.

\frac{3 ^{4} \cdot x}{lo g _{5} ( 1 2 5 )} = lo g (100) \Leftrightarrow \frac{3 ^{4} \cdot x}{lo g _{5} ( 5 ^{3} )} = lo g (1 0^{2})

Again, we can use the the inverse properties of logarithms to simplify the equation.

\frac{3 ^{4} \cdot x}{lo g _{5} ( 5 ^{3} )} = lo g (1 0^{2}) \Leftrightarrow \frac{3 ^{4} \cdot x}{3} = 2

We now need to isolate

x

on the left-hand side to find the solution.

\frac{3 ^{4} \cdot x}{3} = 2

SimpQuotSimplify quotient

3^{3} \cdot x = 2

DivEqn

LHS / 3^{3} = RHS / 3^{3}

x = \frac{2}{3 ^{3}}

CalcPowCalculate power

x = \frac{2}{2 7}

We have solved the equation, and its solution is

\frac{2}{2 7} .

Using Properties of Logarithms

Properties of Logarithms

Product Property of Logarithms

Rule

Quotient Property of Logarithms

Rule

Power Property of Logarithms

Rule

Example

Simplify the expression

Change of Base Formula

Solving an Exponential Equation using Logarithms

1

2

3

Example

Solve the equation using the common logarithm

Inverse Properties of Logarithms

Example

Solve the equation using the inverse properties of logarithms