Logarithm - Example – Application Of Change Of Base Property

Topic: Logarithm - Example – Application of Change of Base Property

In this lecture, we will discuss the application of the Change of Base Property in logarithms.
The Change of Base Property allows us to compute logarithms with different bases by using logarithms with a known base.
We will also go through some examples to understand the concept better.

Logarithm Review

Logarithm is the inverse operation of exponentiation.
It helps us solve exponential equations and perform calculations involving very large or very small numbers.
The general form of a logarithm is given by: logₐ(b) = c, where a is the base, b is the argument, and c is the exponent.
Logarithms have certain properties, such as the Product Rule, Quotient Rule, and Power Rule, that help simplify calculations.

Change of Base Property

The Change of Base Property allows us to calculate logarithms with any base by using logarithms with a known base.
If we have a logarithm with base a, and we want to evaluate it with a base b, we can use the formula: logₐ(b) = logₓ(b) / logₓ(a), where x can be any base.

Applications of Change of Base Property

The Change of Base Property is particularly useful when we want to calculate logarithms with bases that are not readily available on our calculator.
It allows us to convert logarithms with any base into logarithms with bases that our calculator can handle (usually base 10 or base e).

Example 1

Let’s say we want to evaluate log₂(5). However, our calculator only has buttons for logarithms with base 10.
We can use the Change of Base Property to convert log₂(5) to a logarithm with base 10: log₂(5) = log₁₀(5) / log₁₀(2)

Example 2

Suppose we need to calculate log₄(7). However, our calculator only has buttons for logarithms with base e.
We can use the Change of Base Property to convert log₄(7) to a logarithm with base e: log₄(7) = ln(7) / ln(4)

Example 3

Let’s consider an example where we want to evaluate log₅(2). Our calculator does not have a specific button for logarithms with base 5.
We can still use the Change of Base Property to calculate it: log₅(2) = log₃(2) / log₃(5)

Example 4

Suppose we need to find log₉(8), but our calculator only supports logarithms with base 10.
We can utilize the Change of Base Property to calculate log₉(8) as follows: log₉(8) = log₁₀(8) / log₁₀(9)

Example 5

Let’s consider a scenario where we want to evaluate logₖ(1), where k is any positive number.
We can use the Change of Base Property to compute it: logₖ(1) = log₁₀(1) / log₁₀(k) = 0 / log₁₀(k) = 0

Recap

In this lecture, we discussed the Change of Base Property in logarithms.
The Change of Base Property allows us to evaluate logarithms with different bases using logarithms with a known base.
We looked at various examples to understand how to apply the Change of Base Property effectively. Sure! Here are slides 11 to 20 for your lecture on “Logarithm - Example – Application of Change of Base Property”:

Example 6: Evaluate log₇(49) using the Change of Base Property: log₇(49) = log₁₀(49) / log₁₀(7)
Simplifying further, we get: log₁₀(49) / log₁₀(7) = 2 / log₁₀(7)
Therefore, log₇(49) = 2 / log₁₀(7)

Example 7: Calculate log₄(1/16) using the Change of Base Property: log₄(1/16) = log₁₀(1/16) / log₁₀(4)
Simplifying further, we have: log₁₀(1/16) / log₁₀(4) = -2 / log₁₀(4)
Hence, log₄(1/16) = -2 / log₁₀(4)

Example 8: Evaluate log₇(1/49) using the Change of Base Property: log₇(1/49) = log₁₀(1/49) / log₁₀(7)
Simplifying the expression, we get: log₁₀(1/49) / log₁₀(7) = -2 / log₁₀(7)
Therefore, log₇(1/49) = -2 / log₁₀(7)

Example 9: Let’s find log₃(27) by applying the Change of Base Property: log₃(27) = log₁₀(27) / log₁₀(3)
Simplifying the equation, we have: log₁₀(27) / log₁₀(3) = 3 / log₁₀(3)
Hence, log₃(27) = 3 / log₁₀(3)

Example 10: Calculate log₅(0.2) using the Change of Base Property: log₅(0.2) = log₁₀(0.2) / log₁₀(5)
Simplifying further, we get: log₁₀(0.2) / log₁₀(5) ≈ -0.69897 / 0.69897
Therefore, log₅(0.2) ≈ -1

Example 11: Evaluate log₇(7³) using the Change of Base Property: log₇(7³) = log₁₀(7³) / log₁₀(7)
Simplifying the expression, we have: log₁₀(7³) / log₁₀(7) = 3 / log₁₀(7)
Hence, log₇(7³) = 3 / log₁₀(7)

Example 12: Calculate log₂(8) using the Change of Base Property: log₂(8) = log₁₀(8) / log₁₀(2)
Simplifying the equation, we get: log₁₀(8) / log₁₀(2) = 3 / log₁₀(2)
Therefore, log₂(8) = 3 / log₁₀(2)

Example 13: Let’s find log₂(0.125) by applying the Change of Base Property: log₂(0.125) = log₁₀(0.125) / log₁₀(2)
Simplifying the equation, we have: log₁₀(0.125) / log₁₀(2) ≈ -0.90309 / 0.30103
Hence, log₂(0.125) ≈ -3

Example 14: Evaluate log₄(16) using the Change of Base Property: log₄(16) = log₁₀(16) / log₁₀(4)
Simplifying further, we get: log₁₀(16) / log₁₀(4) = 2 / log₁₀(4)
Therefore, log₄(16) = 2 / log₁₀(4)

Example 15: Calculate log₁₀(1000) using the Change of Base Property: log₁₀(1000) = log₅(1000) / log₅(10)
Simplifying the expression, we have: log₅(1000) / log₅(10) ≈ 3 / 0.69897
Therefore, log₁₀(1000) ≈ 4.29003

I hope these slides are helpful for your lecture on the topic of the Change of Base Property in logarithms. Let me know if there’s anything else I can assist you with!

Example 16: Let’s find the value of log₃(81) using the Change of Base Property: log₃(81) = log₁₀(81) / log₁₀(3)
Simplifying further, we have: log₁₀(81) / log₁₀(3) = 4 / log₁₀(3)
Therefore, log₃(81) = 4 / log₁₀(3)

Example 17: Calculate log₅(25) using the Change of Base Property: log₅(25) = log₁₀(25) / log₁₀(5)
Simplifying the expression, we get: log₁₀(25) / log₁₀(5) = 2 / log₁₀(5)
Hence, log₅(25) = 2 / log₁₀(5)

Example 18: Evaluate log₂(16) using the Change of Base Property: log₂(16) = log₁₀(16) / log₁₀(2)
Simplifying further, we have: log₁₀(16) / log₁₀(2) = 4 / log₁₀(2)
Therefore, log₂(16) = 4 / log₁₀(2)

Example 19: Let’s find log₃(243) by applying the Change of Base Property: log₃(243) = log₁₀(243) / log₁₀(3)
Simplifying the equation, we have: log₁₀(243) / log₁₀(3) = 5 / log₁₀(3)
Hence, log₃(243) = 5 / log₁₀(3)

Example 20: Calculate log₄(64) using the Change of Base Property: log₄(64) = log₁₀(64) / log₁₀(4)
Simplifying further, we have: log₁₀(64) / log₁₀(4) = 3 / log₁₀(4)
Therefore, log₄(64) = 3 / log₁₀(4)

Example 21: Evaluate log₉(81) using the Change of Base Property: log₉(81) = log₁₀(81) / log₁₀(9)
Simplifying the expression, we have: log₁₀(81) / log₁₀(9) = 2 / log₁₀(9)
Hence, log₉(81) = 2 / log₁₀(9)

Example 22: Let’s find log₄(256) by applying the Change of Base Property: log₄(256) = log₁₀(256) / log₁₀(4)
Simplifying the equation, we have: log₁₀(256) / log₁₀(4) = 4 / log₁₀(4)
Therefore, log₄(256) = 4 / log₁₀(4)

Example 23: Calculate log₂(32) using the Change of Base Property: log₂(32) = log₁₀(32) / log₁₀(2)
Simplifying further, we have: log₁₀(32) / log₁₀(2) = 5 / log₁₀(2)
Hence, log₂(32) = 5 / log₁₀(2)

Example 24: Evaluate log₅(125) using the Change of Base Property: log₅(125) = log₁₀(125) / log₁₀(5)
Simplifying the expression, we have: log₁₀(125) / log₁₀(5) = 3 / log₁₀(5)
Therefore, log₅(125) = 3 / log₁₀(5)

Example 25: Let’s find log₄(1) by applying the Change of Base Property: log₄(1) = log₁₀(1) / log₁₀(4)
Simplifying the equation, we have: log₁₀(1) / log₁₀(4) = 0 / log₁₀(4)
Hence, log₄(1) = 0 / log₁₀(4)