## 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 $x,$ $\log_b x,$ is the power to which you need to raise $b$ in order to get $x.$ Symbolically,
 $\log_b x = y$ means $b_y = x.$ Logarithmic form Exponential form
Notes
1. $\log_b x$ is only defined if $b$ and $x$ are both positive, and $b \neq 1.$
2. $\log_10 x$ is called the common logarithm of $x,$ and is sometimes written as $\log x.$
3. $\log_e x$ 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 $10^3 = 1\,000$ $4^2 = 16$ $3^3 = 27$ $5^1 = 5$ $7^0 = 1$ $4^{-2} = 1/16$ $25^{1/2} = 5$ Logarithmic Form $\log_{10} 1\,000 = 3$ $\log_4 16 = 2$ $\log_3 27 = 3$ $\log_5 5 = 1$ $\log_7 1 = 0$ $\log_4 (1/16) = -2$ $\log_{25} 5 = 1/2$

Here are some for you to try

Exponential Form$10^2 = 100$$3^{-2} = 1/9$
Logarithmic Form
 $\log$ =

 $\log$ =

 Exponential Form ^ = ^ = Logarithmic Form $\log_3 1 =0$ $\log_5 (1/125) = -3$

Example 1 Calculating Logarithms by Hand

 (a) $\log_2 8$ = Power to which you need to raise $2$ in order to get $8$ = 3       Since $2^3 = 8$ (b) $\log_4 1$ = Power to which you need to raise $4$ in order to get $1$ = $0$       Since $4^0 = 1$ (c) $\log_{10} 10\,000$ = Power to which you need to raise $10$ in order to get $10\,000$ = $4$       Since $10^4 = 10\,000$ (d) $\log_{10} 1/100$ = Power to which you need to raise $10$ in order to get $1/100$ = $-2$       Since $10^{-2} = 1/100$ (e) $\log_3 27$ = (f) $\log_9 3$ = (g) $\log_3 (1/81)$ =

Algebraic Properties of Logarithms

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

Identity
Example
(a) $\log_a(xy) = \log_ax + \log_ay$ $\log_216 = \log_28 + \log_22$
(b) $\log_a\left( \frac{x}{y} \right) = \log_ax - \log_a y$ $\log_2 \left( \frac{5}{3} \right) = \log_25 - \log_23$
(c) $\log_a(x^r) = r \log_ax$ $\log_2(6^5) = 5 \log_26$
 (d) $\log_aa=1$ $\log_a1=0$
 $\log_22 = 1$ $\log_31 = 0$
(e) $\log_a\left( \frac{1}{x} \right) = -\log_ax$ $\log_2\left( \frac{1}{3} \right) = -\log_23$
(f) $\log_ax = \frac{\log x}{\log a} = \frac{\ln x}{\ln a}$ $\log_25 = \frac{\log 5}{\log 2}\approx 2.3219$

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) $\log 32$ $\log 2 ^5 = 5 \log 2 = 5a$ (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 $\log_axy = \log_ax + \log_ay$ ?
A Let $s = \log_ax,$ and $t = \log_a y.$ In exponential form, these equations say that
$a^s = x$ and $a^t = y.$
Multiplying these two equations together gives
$a^s a^t = xy,$
that is,
$a^{s+t} = xy.$
Rewriting this in logarithmic form gives
$\log_a (xy) = s + t = \log_a x + \log_a y$
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 $\log_a(x^r) = r \log_ax$ ?
A Let $t = \log_a x.$ Writing this in exponential form gives
$a^t = x.$
Raising this equation to the $r^{th}$ power gives
$a^{rt} = x^{r}.$
Rewriting in logarithmic form gives
$\log_a (x^r) = rt = r \log_a x,$
as claimed.

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

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

Q Why is

$\log_ax = \frac{\log x}{\log a} = \frac{\ln x}{\ln a}$?
A Let $s = \log_a x.$ In exponential form, this says that
$a^s = x.$
Take the logarithm with base b of both sides, getting
$\log_b a^s = \log_b x,$
then use identity (c):
$s \log_b a = \log b x,$
so
$s = \frac{\log_bx}{\log_ba}$

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) $4^{-x^2} = 1/64.$
(b) $5 (1.1^{2x+3}) = 200$

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

(a) Write the given equation in logarithmic form:

 $4^{-x^2} = 1/64$ Exponential Form $\log_4(1/64) = -x^2$ Logarithmic Form Thus, $-x^2 = \log_4(1/64) = -3$ giving $x = ±3^{1/2}.$

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

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

Last Updated:October, 1999