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 
. 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
Examples
Example 1
Calculate log 2 32.
Solution 1
We see that
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
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 b ∈ 
is arbitrary then
log a ( a b ) = b .
If a > 0, b > 0, and c > 0 then
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
for the unknown x .
Solution 4
We take the natural logarithm of both sides:
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
Example 5
Simplify the expression
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
You Try It : What does 
mean (in terms of the natural logarithm function)?
Example 6
Simplify the expression (log 4 9) · (log 9 16).
Solution 6
We have
Find practice problems and solutions for these concepts at: Transcendental Functions Practice Test.
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