**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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