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

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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 formExponential form
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.

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

Exponential Form103 = 1,00042 = 1633 = 2751 = 570 = 14-2 = 1/16251/2 = 5
Logarithmic Formlog101,000 = 3log416 = 2log327 = 3log55 = 1log71 = 0log4(1/16) =-2log255 = 1/2

Here are some for you to try

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



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

Example 1 Calculating Logarithms by Hand

Algebraic Properties of Logarithms

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

(a) loga(xy)= logax + logay
log216 = log28 + log22
(b) loga

= logax -logay

= log25 - log23
(c) loga(xr)= r logax
log2(65) = 5 log26
(d) logaa =1
         loga1 =0
log31 =0
(e) loga

= -logax

= -log23
(f) logax =
log x

log a
ln x

ln a
log25 =
log 5

log 2

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.

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,
s =


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.

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

(a) Write the given equation in logarithmic form:

        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.

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