Logarithmic Function Identity

Rewriting Products and Powers with Logarithms

Logarithmic identities turn multiplication into addition, division into subtraction, and powers into coefficients. Use them to split one logarithm into several terms or combine several terms into one logarithm.

Logarithms undo exponentiation. If $b^x = a$ , then $\log_b a = x$ . Every example below uses a positive logarithm argument and a base that satisfies $b>0$ and $b\neq1$ .

Basic Logarithmic Identities

When a term contains a product, a quotient, or a power, one of these rules splits it into separate logarithms first. After that the expression can often be evaluated mentally.

The identities below are the ones you reach for first.

Product Identity

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

The logarithm of a product equals the sum of the logarithms of each number.

Example:

\begin{aligned} \log_2(8 \times 4) &= {\log_2 8} + {\log_2 4} \\ &= 3 + 2 \\ &= 5 \end{aligned}

Quotient Identity

\log_b\left(\frac{M}{N}\right) = {\log_b M} - {\log_b N}

The logarithm of a quotient equals the difference between the logarithm of the numerator and the logarithm of the denominator.

Example:

\begin{aligned} \log_3\left(\frac{81}{9}\right) &= {\log_3 81} - {\log_3 9} \\ &= 4 - 2 \\ &= 2 \end{aligned}

Power Identity

\log_b(M^p) = p \cdot {\log_b M}

The logarithm of a number raised to a power equals the power multiplied by the logarithm of that number.

Example:

\log_2(4^3) = 3 \cdot {\log_2 4} = 3 \times 2 = 6

Special Logarithmic Identities

Each rule below names the case it applies to before it shows the rewrite. Always check the base and the argument before you use one of them.

Some identities hold only for a particular base or a particular argument.

Change of Base

\log_b M = \frac{\log_a M}{\log_a b}

Use this identity to rewrite a logarithm in a base that is available or easier to evaluate.

Example:

Because $9=3^2$ , the numerator is exactly twice the denominator:

\log_3 9 = \frac{\log_{10} 9}{\log_{10} 3} = \frac{2\log_{10}3}{\log_{10}3} = 2

Equality Identity

If $\log_b M = {\log_b N}$ , then $M = N$

Two numbers that have the same logarithmic value (with the same base) must be the same number. For example, $\log_2 8 = \log_2 2^3$ and both sides equal $3$ , so the arguments $8$ and $2^3$ are equal. The rule works because each value of a logarithm with a fixed base comes from exactly one argument.

Comparing Two Logarithms

This identity links the sign of a logarithm to where its argument sits relative to the base, which is what you need when a logarithm appears inside an inequality.

If $b > 1$ and $\log_b M < {\log_b N}$ , then $M < N$
If $0 < b < 1$ and $\log_b M < {\log_b N}$ , then $M > N$

Applications of Logarithmic Identities

Here an identity meets an algebraic expression, so the expression has to reach a suitable form first. Work step by step and note which identity you use at each step.

The identities earn their place when an expression must be simplified before it can be solved.

Simplifying Expressions

Simplify $\log_2 32 + {\log_2 8} - {\log_2 4}$

Solution:

\log_2 32 + {\log_2 8} - {\log_2 4}

Solving Equations

Find the value of $x$ if $\log_3 x + {\log_3 9} = {\log_3 81}$

Solution:

\log_3 x + {\log_3 9} = {\log_3 81}

Applications in Models

In a model the unknown quantity sits in the exponent, and a logarithm brings it down. The examples below show which identity each one needs.

Logarithmic identities also appear when a model contains exponential growth or decay.

A Logarithmic Magnitude Scale

A simplified magnitude scale can compare a measured seismic-wave amplitude $A$ with a reference amplitude $A_0$ :

M = \log\left(\frac{A}{A_0}\right)

Where:

$M$ is the magnitude in this simplified model
$A$ is the measured amplitude
$A_0$ is the reference amplitude

Example: Suppose a measured amplitude is $10^7$ times the reference amplitude. What magnitude does the simplified model give?

Solution:

M = \log\left(\frac{A}{A_0}\right)

The amplitude ratio therefore corresponds to magnitude $7$ in this simplified model. Actual earthquake-magnitude calculations also account for the instrument, wave type, and source distance.

Battery Charging

In an idealized first-order charging model, the time needed to reach a fraction of the maximum capacity is:

t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)

Where:

$t$ = charging time (in minutes)
$k$ = charging constant
$C$ = desired capacity
$C_0$ = maximum capacity

Example: Determine the time required to charge a battery from empty to $90\%$ full. Assume $k = 0.02$ .

Solution:

t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)

Within this idealized model, reaching $90\%$ takes approximately $115 \text{ minutes}$ . Real charging systems can change their rate near full capacity.

Car Price Depreciation

If the instantaneous depreciation rate is proportional to a car's current value, an exponential model has the form:

H(t) = ce^{kt}

In this model, $H(t)$ is the car's value at time $t$ .

Example: A new car costs $200 \text{ million rupiah}$ and its modeled value after $5 \text{ years}$ is $100 \text{ million rupiah}$ . Assuming the same exponential depreciation rate continues, determine the modeled value after $10 \text{ years}$ .

Solution:

Measure $H(t)$ in millions of rupiah and $t$ in years. At purchase:

\begin{aligned} &H(0)=200 \\ &200=ce^0=c \end{aligned}

From these results, the car's price at any time $t$ is:

H(t) = 200e^{\frac{\ln(1/2)}{5}t} \approx 200e^{-0.1386t}

The car's price after $10 \text{ years}$ of use is:

\begin{aligned} H(10) &= 200e^{2\ln(1/2)} \\ &= 200\left(\frac{1}{2}\right)^2 \\ &= 50 \text{ million rupiah} \end{aligned}

Exercises

The three problems each need a different rewrite, so write down which rule you need before you calculate. Then compare your route with the route the solution takes.

Problem 1

Simplify: $\log_5 125 + {\log_5 25} - {\log_5 5}$

Problem 2

If $\log_2 x = 3$ and $\log_2 y = 5$ , find the value of $\log_2(xy)$

Problem 3

Find the value of $x$ if $\log_4(x-1) = {\log_4 16} - {\log_4 2}$

Worked Solutions

Answer $1$

Each number is a power of $5$ . Evaluate the three logarithms before combining them:

125=5^3,\quad25=5^2,\quad5=5^1

The simplified value is $4$ .

Answer $2$

Use the product identity because the target contains $xy$ :

\log_2(xy) = {\log_2 x} + {\log_2 y}

$\log_2(xy)=8$ .

Answer $3$

Use the quotient identity on the right-hand side, then use equality of logarithms with the same base:

\log_4(x-1) = {\log_4 16} - {\log_4 2}

The logarithm requires $x-1>0$ . Since $9-1=8>0$ , the solution is valid.

Rewriting Products and Powers with Logarithms

Logarithms undo exponentiation. If $b^x = a$ , then $\log_b a = x$ . Every example below uses a positive logarithm argument and a base that satisfies $b>0$ and $b\neq1$ .

Basic Logarithmic Identities

When a term contains a product, a quotient, or a power, one of these rules splits it into separate logarithms first. After that the expression can often be evaluated mentally.

The identities below are the ones you reach for first.

Product Identity

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

The logarithm of a product equals the sum of the logarithms of each number.

Example:

\begin{aligned} \log_2(8 \times 4) &= {\log_2 8} + {\log_2 4} \\ &= 3 + 2 \\ &= 5 \end{aligned}

Quotient Identity

\log_b\left(\frac{M}{N}\right) = {\log_b M} - {\log_b N}

The logarithm of a quotient equals the difference between the logarithm of the numerator and the logarithm of the denominator.

Example:

\begin{aligned} \log_3\left(\frac{81}{9}\right) &= {\log_3 81} - {\log_3 9} \\ &= 4 - 2 \\ &= 2 \end{aligned}

Power Identity

\log_b(M^p) = p \cdot {\log_b M}

The logarithm of a number raised to a power equals the power multiplied by the logarithm of that number.

Example:

\log_2(4^3) = 3 \cdot {\log_2 4} = 3 \times 2 = 6

Special Logarithmic Identities

Each rule below names the case it applies to before it shows the rewrite. Always check the base and the argument before you use one of them.

Some identities hold only for a particular base or a particular argument.

Change of Base

\log_b M = \frac{\log_a M}{\log_a b}

Use this identity to rewrite a logarithm in a base that is available or easier to evaluate.

Example:

Because $9=3^2$ , the numerator is exactly twice the denominator:

\log_3 9 = \frac{\log_{10} 9}{\log_{10} 3} = \frac{2\log_{10}3}{\log_{10}3} = 2

Equality Identity

If $\log_b M = {\log_b N}$ , then $M = N$

Comparing Two Logarithms

This identity links the sign of a logarithm to where its argument sits relative to the base, which is what you need when a logarithm appears inside an inequality.

If $b > 1$ and $\log_b M < {\log_b N}$ , then $M < N$
If $0 < b < 1$ and $\log_b M < {\log_b N}$ , then $M > N$

Applications of Logarithmic Identities

Here an identity meets an algebraic expression, so the expression has to reach a suitable form first. Work step by step and note which identity you use at each step.

The identities earn their place when an expression must be simplified before it can be solved.

Simplifying Expressions

Simplify $\log_2 32 + {\log_2 8} - {\log_2 4}$

Solution:

\log_2 32 + {\log_2 8} - {\log_2 4}

Solving Equations

Find the value of $x$ if $\log_3 x + {\log_3 9} = {\log_3 81}$

Solution:

\log_3 x + {\log_3 9} = {\log_3 81}

Applications in Models

In a model the unknown quantity sits in the exponent, and a logarithm brings it down. The examples below show which identity each one needs.

Logarithmic identities also appear when a model contains exponential growth or decay.

A Logarithmic Magnitude Scale

A simplified magnitude scale can compare a measured seismic-wave amplitude $A$ with a reference amplitude $A_0$ :

M = \log\left(\frac{A}{A_0}\right)

Where:

$M$ is the magnitude in this simplified model
$A$ is the measured amplitude
$A_0$ is the reference amplitude

Example: Suppose a measured amplitude is $10^7$ times the reference amplitude. What magnitude does the simplified model give?

Solution:

M = \log\left(\frac{A}{A_0}\right)

The amplitude ratio therefore corresponds to magnitude $7$ in this simplified model. Actual earthquake-magnitude calculations also account for the instrument, wave type, and source distance.

Battery Charging

In an idealized first-order charging model, the time needed to reach a fraction of the maximum capacity is:

t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)

Where:

$t$ = charging time (in minutes)
$k$ = charging constant
$C$ = desired capacity
$C_0$ = maximum capacity

Example: Determine the time required to charge a battery from empty to $90\%$ full. Assume $k = 0.02$ .

Solution:

t = -\frac{1}{k}\ln\left(1 - \frac{C}{C_0}\right)

Within this idealized model, reaching $90\%$ takes approximately $115 \text{ minutes}$ . Real charging systems can change their rate near full capacity.

Car Price Depreciation

If the instantaneous depreciation rate is proportional to a car's current value, an exponential model has the form:

H(t) = ce^{kt}

In this model, $H(t)$ is the car's value at time $t$ .

Solution:

Measure $H(t)$ in millions of rupiah and $t$ in years. At purchase:

\begin{aligned} &H(0)=200 \\ &200=ce^0=c \end{aligned}

From these results, the car's price at any time $t$ is:

H(t) = 200e^{\frac{\ln(1/2)}{5}t} \approx 200e^{-0.1386t}

The car's price after $10 \text{ years}$ of use is:

\begin{aligned} H(10) &= 200e^{2\ln(1/2)} \\ &= 200\left(\frac{1}{2}\right)^2 \\ &= 50 \text{ million rupiah} \end{aligned}

Exercises

The three problems each need a different rewrite, so write down which rule you need before you calculate. Then compare your route with the route the solution takes.

Problem 1

Simplify: $\log_5 125 + {\log_5 25} - {\log_5 5}$

Problem 2

If $\log_2 x = 3$ and $\log_2 y = 5$ , find the value of $\log_2(xy)$

Problem 3

Find the value of $x$ if $\log_4(x-1) = {\log_4 16} - {\log_4 2}$

Worked Solutions

Answer $1$

Each number is a power of $5$ . Evaluate the three logarithms before combining them:

125=5^3,\quad25=5^2,\quad5=5^1

The simplified value is $4$ .

Answer $2$

Use the product identity because the target contains $xy$ :

\log_2(xy) = {\log_2 x} + {\log_2 y}

$\log_2(xy)=8$ .

Answer $3$

Use the quotient identity on the right-hand side, then use equality of logarithms with the same base:

\log_4(x-1) = {\log_4 16} - {\log_4 2}

The logarithm requires $x-1>0$ . Since $9-1=8>0$ , the solution is valid.