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)	$\log_3 (1/81)$	=

 
Example 2 Using the Properties of Logarithms

		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.

$a^s = x$ and $a^t = y.$

$a^s a^t = xy,$
that is,

$a^{s+t} = xy.$

$\log_a (xy) = s + t = \log_a x + \log_a y$

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.

$a^t = x.$

$a^{rt} = x^{r}.$

$\log_a (x^r) = rt = r \log_a x,$

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

Q Why is

$\log_ax = \frac{\log x}{\log a} = \frac{\ln x}{\ln a}$?

$a^s = x.$

$\log_b a^s = \log_b x,$

$s \log_b a = \log b x,$

$s = \frac{\log_bx}{\log_ba}$

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

Example 3 Solving for the Exponent

(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.