## Using and DerivingAlgebraic Properties of Logarithms miscellaneous on-line topics for Calculus Applied to the Real World

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Logarithms

We start by reviewing the basic definitions as in Section 2.3 of Calculus Applied to the Real World. If you like, you can also take a look at the topic summary material on logarithms.

Base b Logarithm
The base b logarithm of x, logbx, is the power to which you need to raise b in order to get x. Symbolically,
 logbx = y means by = x. Logarithmic form Exponential form
Notes
1. logbx is only defined if b and x are both positive, and b 1.
2. log10x is called the common logarithm of x, and is sometimes written as log 10.
3. logex is called the natural logarithm of x and is sometimes written as ln x.

Examples
The following table lists some exponential equations and their equivalent logarithmic form.

 Exponential Form 103 = 1,000 42 = 16 33 = 27 51 = 5 70 = 1 4-2 = 1/16 251/2 = 5 Logarithmic Form log101,000 = 3 log416 = 2 log327 = 3 log55 = 1 log71 = 0 log4(1/16) =-2 log255 = 1/2

Here are some for you to try

Exponential Form102 = 1003-2 = 1/9
Logarithmic Form
 log =

 log =

 Exponential Form ^ = ^ = Logarithmic Form log31 =0 log5(1/125) = -3

Example 1 Calculating Logarithms by Hand

 (a) log28 = Power to which you need to raise 2 in order to get 8 = 3       Since 23 = 8 (b) log41 = Power to which you need to raise 4 in order to get 1 = 0       Since 40 = 1 (c) log1010,000 = Power to which you need to raise 10 in order to get 10,000 = 4       Since 104 = 10,000 (d) log101/100 = Power to which you need to raise 10 in order to get 1/100 = -2       Since 10-2 = 1/100 (e) log327 = (f) log93 = (g) log3(1/81) =

Algebraic Properties of Logarithms

The following identities hold for any positive a 1 and any positive numbers x and y.

Identity
Example
 (a) loga(xy) = logax + logay
 log216 = log28 + log22
 (b) loga xy = logax -logay
 log2 53 = log25 - log23
 (c) loga(xr) = r logax
 log2(65) = 5 log26
 (d) logaa = 1 loga1 = 0
 log22 = 1 log31 = 0
 (e) loga 1x = -logax
 log2 13 = -log23
 (f) logax = log x log a = ln x ln a
 log25 = log 5 log 2 2.3219

Example 2 Using the Properties of Logarithms

Let a = log 2, b = log 3, and c = log 5. Write the following in terms of a, b, and c.
Note If any answer you give is not simplified -- for instance, if you say a + a instead of 2a -- it will be marked wrong.

 Answer (a) log 6 log 2 + log 3 = a + b (b) log 15 (c) log 30 log 2 + log 3 + log 5 = a + b + c (d) log 12 (e) log 1.5 log 3 - log 2 = b - a (f) log(1/9) (g) log 32 log 2 = 5 log 2 = 5a (h) log(1/81)

Q Where do the identities come from?
A Roughly speaking, they are restatements in logarithmic form of the laws of exponents.

Q Why is logaxy = logax + logay ?
A Let s = logax and t = logay. In exponential form, these equations say that

as = x and at = y.
Multiplying these two equations together gives
asat = xy,
that is,
as+t = xy.
Rewriting this in logarithmic form gives
loga(xy) = s + t = logax + logay
as claimed.

Here is an intuitive way of thinking about it: Since logs are exponents, this identity expresses the familiar law that the exponent of a product is the sum of the exponents.

The second logarithmic identity is shown in almost the identical way, and we leave it for you for practice.

Q Why is loga(xr) = r logax ?
A Let t = logax. Writing this in exponential form gives

at = x.
Raising this equation to the rth power gives
art = xr.
Rewriting in logarithmic form gives
loga(xr) = rt = rlogax,
as claimed.

Identity (d) we will leave for you to do as practice.

Q Why is loga(1/x) = -logax ?
A This follows from identities (b) and (d) (think about it). <>

Q Why is

 logax = log x log a = ln x ln a ?
A Let s = logax. In exponential form, this says that
as = x.
Take the logarithm with base b of both sides, getting
logbas = logbx,
then use identity (c):
slogba = logbx,
so
 s = logbxlogba

Since logarithms are exponents, we can use them to solve equations where the unknown is in the exponent.

Example 3 Solving for the Exponent

Solve the following equations for x.

(a) 4-x2 = 1/64.
(b) 5 (1.12x+3) = 200

Solution We can solve both of these equations by translating from exponential form to logarithmic form.

(a) Write the given equation in logarithmic form:

tr>
 4-x2 = 1/64 Exponential Form log4(1/64) = -x2 Logarithmic Form Thus, -x2 = log4(1/64) = -3 giving x = 31/2.

(b) Before converting to logarithmic form, first divide both sides of the equation by 5:
 5 (1.12x+3) = 200 1.12x+3 = 40 Exponential Form log1.140 = 2x+3 Logarithmic Form This gives 2x + 3 = ln 40/ln 1.1 38.7039, Identity (e) so that x 17.8520.

You can now either go on and try the exericses in the exercise set for this topic.

Last Updated:October, 1999
Copyright © 1999 StefanWaner and Steven R. Costenoble