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4.4 Laws Of Logarithms

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Objectives

► Laws of Logarithms

► Expanding and Combining Logarithmic Expressions

► Change of Base Formula

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Laws of Logarithms

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Laws of LogarithmsSince logarithms are exponents, the Laws of Exponents give rise to the Laws of Logarithms.

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Example 1 – Using the Laws of Logarithms to Evaluate Expressions

Evaluate each expression.

(a) log4 2 + log4 32

(b) log2 80 – log2 5

(c) log 8

Solution:(a) log4 2 + log4 32 = log4(2 32)

= log4 64 = 3

Law 1

Because 64 = 43

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Example 1 – Solution(b) log2 80 – log2 5 = log2

= log216 = 4

(c) log 8 = log (8–1/3)

= log

–0.301

Because 16 = 24

Law 2

cont’d

Law 3

Property of negative exponents

Calculator

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Expanding and Combining Logarithmic Expressions

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Expanding and Combining Logarithmic Expressions

The Laws of Logarithms allow us to write the logarithm of aproduct or a quotient as the sum or difference of logarithms.

This process, called expanding a logarithmic expression, is illustrated in the next example.

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Example 2 – Expanding Logarithmic Expressions

Use the Laws of Logarithms to expand each expression.

(a) log2(6x) (b) log5(x3y6) (c) ln

Solution:(a) log2(6x) = log2 6 + log2 x

(b) log5(x3y6) = log5 x3 + log5 y6

= 3 log5 x + 6 log5 y

Law 1

Law 1

Law 3

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Example 2 – Solution(c) = ln(ab) –

= ln a + ln b – ln c1/3

= ln a + ln b – ln c

Law 2

Law 1

Law 3

cont’d

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Expanding and Combining Logarithmic Expressions

The Laws of Logarithms also allow us to reverse the process of expanding that was done in Example 2.

That is, we can write sums and differences of logarithms as a single logarithm.

This process, called combining logarithmic expressions, is illustrated in the next example.

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Example 3 – Combining Logarithmic Expressions

Combine 3 log x + log (x + 1) into a single logarithm.

Solution:3log x + log(x + 1) = log x3 + log(x + 1)1/2

= log(x3(x + 1)1/2)

Law 3

Law 1

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Expanding and Combining Logarithmic Expressions

Logarithmic functions are used to model a variety of situations involving human behavior.

One such behavior is how quickly we forget things we have learned.

For example, if you learn algebra at a certain performance level (say, 90% on a test) and then don’t use algebra for a while, how much will you retain after a week, a month, or a year?

Hermann Ebbinghaus (1850–1909) studied this phenomenon and formulated the law described in the next example.

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Example 5 – The Law of ForgettingIf a task is learned at a performance level P0, then after a time interval t the performance level P satisfies the equation

log P = log P0 – c log(t + 1),

where c is a constant that depends on the type of task and t is measured in months.

(a) Solve the equation for P.

(b) If your score on a history test is 90, what score wouldyou expect to get on a similar test after two months?After a year? (Assume that c = 0.2.)

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Example 5 – Solution(a) We first combine the right-hand side.

log P = log P0 – c log(t + 1)

log P = log P0 – log(t + 1)c

log P =

P =

Given equation

Law 3

Law 2

Because log is one-to-one

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Example 5 – Solution(b) Here P0 = 90, c = 0.2, and t is measured in months.

In two months: t = 2 and

In one year: t = 12 and

Therefore, your expected scores after two months and one year are 72 and 54, respectively.

cont’d

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Change of Base Formula

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Change of Base FormulaFor some purposes we find it useful to change from logarithms in one base to logarithms in another base.

Suppose we are given logax and want to find logbx.

Lety = logb x

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Change of Base FormulaWe write this in exponential form and take the logarithm, with base a, of each side.

by = x

loga(by) = logax

y loga b = logax

Exponential form

Take loga of each side

Law 3

Divide by logab

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Change of Base FormulaThis proves the following formula.

In particular, if we put x = a, then logaa = 1, and this formula becomes

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Example 6 – Evaluating Logarithms with the Change of Base Formula

Use the Change of Base Formula and common or natural logarithms to evaluate each logarithm, correct to five decimal places.

(a) log85 (b) log920

Solution:(a) We use the Change of Base Formula with b = 8 and

a = 10:

log8 5 = 0.77398

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Example 6 – Solution(b) We use the Change of Base Formula with b = 9 and

a = e:

log9 20 = 1.36342

cont’d