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Logarithms with Arbitrary Bases Help

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By — McGraw-Hill Professional
Updated on Aug 31, 2011

Introduction to Logarithms with Arbitrary Bases

If you review the first few paragraphs of Section 1, you will find an intuitively appealing definition of the logarithm to the base 2:

log 2 x is the power to which you need to raise 2 to obtain x .

With this intuitive notion we readily see that

log 2 16 = “the power to which we raise 2 to obtain 16” = 4

and

log 2 (1/4) = “the power to which we raise 2 to obtain 1/4” = −2.

However this intuitive approach does not work so well if we want to take log π 5 or Transcendental Functions 6.3 Exponentials with Arbitrary Bases . Therefore we will give a new definition of the logarithm to any base a > 0 which in simple cases coincides with the intuitive notion of logarithm.

If a > 0 and b > 0 then

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Examples

Example 1

Calculate log 2 32.

Solution 1

We see that

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Notice that, in this example, the new definition of log 2 32 agrees with the intuitive notion just discussed.

Example 2

Express ln x as the logarithm to some base.

Solution 2

If x > 0 then

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Thus we see that the natural logarithm ln x is precisely the same as log e x .

Math Note : In mathematics, it is common to write ln x rather than log e x .

You Try It : Calculate log 3 27 + log 5 (1/25) − log 2 8.

We will be able to do calculations much more easily if we learn some simple properties of logarithms and exponentials.

If a > 0 and b > 0 then

a (log a b ) = b .

If a > 0 and bTranscendental Functions 6.3 Exponentials with Arbitrary Bases is arbitrary then

log a ( a b ) = b .

If a > 0, b > 0, and c > 0 then

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

We next give several examples to familiarize you with logarithmic and exponential operations.

Example 3

Simplify the expression

log 3 81 − 5 · log 2 8 − 3 · ln( e 4 ).

Solution 3

The expression equals

log 3 (3 4 ) − 5·log 2 (2 3 )−3·ln e 4 = 4·log 3 3 − 5·[3·log 2 2] − 3·[4·ln e ]

= 4·1 − 5·3·1 − 3·4·1 = −23.

You Try It : What does log 3 5 mean in terms of natural logarithms?

Examples

Example 4

Solve the equation

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

for the unknown x .

Solution 4

We take the natural logarithm of both sides:

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Applying the rules for logarithms we obtain

ln(5 x ) + ln(2 3 x ) = ln 4 − ln(7 x )

or

x · ln 5 + 3 x · ln 2 = ln 4 − x · ln 7.

Gathering together all the terms involving x yields

x · [ln 5 + 3 · ln 2 + ln 7] = ln 4

or

x · [ln(5 · 2 3 · 7)] = ln 4.

Solving for x gives

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Example 5

Simplify the expression

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Solution 5

The numerator of B equals

log 7 (3 5 ) − log 7 (16 1/4 ) = log 7 243 − log 7 2 = log 7 (243/2).

Similarly, the denominator can be rewritten as

log 7 5 3 + log 7 (32 1/5 ) = log 7 125 + log 7 2 = log 7 (125 · 2) = log 7 250.

Putting these two results together, we find that

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

You Try It : What does Transcendental Functions 6.3 Exponentials with Arbitrary Bases mean (in terms of the natural logarithm function)?

Example 6

Simplify the expression (log 4 9) · (log 9 16).

Solution 6

We have

Transcendental Functions 6.3 Exponentials with Arbitrary Bases

Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.

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