Mastering Operations with Logarithms: Key Concepts Explained

Property	Reason
$lo g_{b} b = 1$	A number raised to the power of $1$ is equal to itself.
$lo g_{b} 1 = 0$	A number raised to the power of $0$ is equal to $1 .$

Property

Reason

lo g_{b} b = 1

A number raised to the power of

1

is equal to itself.

lo g_{b} 1 = 0

A number raised to the power of

0

is equal to

1 .

$lo g_{b} a = c \Leftrightarrow a = b^{c}$
$a = b^{c} Substitute a = b^{l o g_{b} a}$
$lo g_{b} a = c Substitute lo g_{b} b^{c} = c$

lo g_{b} a = c \Leftrightarrow a = b^{c}

a = b^{c} Substitute a = b^{l o g_{b} a}

lo g_{b} a = c Substitute lo g_{b} b^{c} = c

Definition	$lo g_{b} a = c \Leftrightarrow a = b^{c}$
Identity Derived From the Definition	$lo g_{b} b = 1$
Identity Derived From the Definition	$lo g_{b} 1 = 0$
Product Property of Logarithms	$lo g_{b} m n = lo g_{b} m + lo g_{b} n$
Quotient Property of Logarithms	$lo g_{b} \frac{m}{n} = lo g_{b} m - lo g_{b} n$
Power Property of Logarithms	$lo g_{b} m^{n} = n lo g_{b} m$

Definition

lo g_{b} a = c \Leftrightarrow a = b^{c}

Identity Derived From the Definition

lo g_{b} b = 1

Identity Derived From the Definition

lo g_{b} 1 = 0

Product Property of Logarithms

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

Quotient Property of Logarithms

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

Power Property of Logarithms

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

Order the logarithms from least to greatest value.

lo g_{5} 23, lo g_{6} 38, lo g_{7} 8, lo g_{2} 10

To order the given logarithms from least to greatest value, we will find lower and upper boundaries for the numerical value of each expression. Let's start with log_5 23. Recall that 5^1=5 and that 5^2=25. With this information, we can find the values of log_5 5 and log_5 25. cc 5^1= 5 & 5^2= 25 ⇕ & ⇕ log_5 5 = 1 & log_5 25= 2 Because 23 is between 5 and 25, we can state that log_5 23 is between log_5 5 and log_5 25. This means that log_5 23 is greater than 1 and less than 2. 13 Let's use a table to summarize the obtained inequalities.

Inequality	Comment
1	log_5 23 is closer to 2 than to 1
2	-
1	log_7 8 is closer to 1 than to 2
log_2 10>3	-

By using the information from the table, we can write the logarithmic expressions in order from least to greatest value. log_7 8, log_5 23, log_6 38, log_2 10

Use the definition of a logarithm to evaluate the logarithms. Write the answers as simplified fractions.

lo g_{125} 25

lo g_{8} 32

lo g_{27} 81

lo g_{4} 128

Let's start by writing 125 and 25, which are the base and the argument of the given logarithm, respectively, as powers with base 5. log_(125) 25 = log_(5^3) 5^2 Therefore, we need to find a number c such that 5^3 raised to the power of c equals 5^2. This can be algebraically expressed by using the definition of a logarithm. log_(5^3) 5^2= c ⇔ ( 5^3)^c= 5^2 We can use the Power of a Power Property to solve the exponential equation. Then, since both sides have the same base, we can use the Property of Equality for Exponential Equations and equate the exponents. Let's do it!

We found that the numerical value of log_(125) 25 is 23.

Similar to Part A, we can start by expressing the base of the logarithm 8 and the argument 32 as powers with base 2. log_8 32 = log_(2^3) 2^5 We need to find a number c such that 2^3 raised to the power of c equals 2^5. Let's use the definition of a logarithm to algebraically express this. log_(2^3) 2^5= c ⇔ ( 2^3)^c= 2^5 We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.

Therefore, the numerical value of log_8 32 is 53.

In a similar manner as before, we can start by expressing the base of the logarithm 27 and the argument 81 as powers with base 3. log_(27) 81 = log_(3^3) 3^4 We need to find a number c such that 3^3 raised to the power of c equals 3^4. Let's use the definition of a logarithm to algebraically express this. log_(3^3) 3^4= c ⇔ ( 3^3)^c= 3^4 We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.

Therefore, the numerical value of log_(27) 81 is 43.

Finally, we can express 4 and 128, which are the base and the argument of the given logarithm, as powers with base 2. log_4 128 = log_(2^2) 2^7 We need to find a number c such that 2^2 raised to the power of c equals 2^7. Let's use the definition of a logarithm to algebraically express this. log_(2^2) 2^7= c ⇔ ( 2^2)^c= 2^7 We can solve the exponential equation by using the Power of a Power Property and the Property of Equality for Exponential Equations.

We found that the numerical value of log_4 128 is 72.

Consider the following logarithmic expression.

lo g x = 5

What is the value of

\frac{1}{x} ?

Before finding the value of 1x, it would be convenient for us to find the value of x. To do so, we will use the definition of a logarithm. Note that we have a common logarithm, which means that the base is 10. Definition:& log_b a= c ⇔ b^c= a Given Logarithm:& log x = 5 ⇔ 10^5= x We found that x is equal to 10^5, or 100 000. With this information, we can find the value of its reciprocal 1x. x=100 000 ⇔ 1/x=1/100 000 Therefore, 1x is equal to 1100 000 or 0.00001.

Properties of Logarithms

Catch-Up and Review

What Is the Inverse Operation of Calculating an Exponent?

Logarithm

Why

Evaluating and Rewriting Logarithmic Expressions

Hint

Solution

Evaluating Logarithms

Rewriting Logarithmic and Exponential Equations

Properties of Logarithms

Product Property of Logarithms

Proof

Quotient Property of Logarithms

Proof

Power Property of Logarithms

Proof

Using Properties to Evaluate Expressions

Hint

Solution

Using Properties to Expand Expressions

Hint

Solution

Using Properties to Condense Expressions

Hint

Solution

Inverse Operations

Properties of Logarithms

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