MAT 124 Logarithmic Functions

What is a Logarithm?

In the early 1600’s, a Scottish mathematician named John Napier introduced a new tool for arithmetical computations: logarithms. This was a revolutionary time-saving device for astronomers at the time. Lengthy calculations that would normally require years could now be done in a few months. This worked because logarithms reduce products and quotients of large numbers into sums and difference of smaller ones.

So what is a logarithm? Well, it’s simply the name given to certain exponents. For example, 11 since 23    0.125 , we write log 0.125  3 . Since 54  625 , we write log 625 4 283 2 5 , and so on.

Note that logarithms are specific to a given base a that is always positive and not equal to 1.

Also, logarithms can only be applied to positive numbers. Thus, loga x is only defined if x  0 .

By convention, we write loge xx ln for the natural base e  2.72 and log10 xx log for the common base 10. These widely-used bases are the ones featured on graphing calculators with the LN and LOG keys.

Note that any logarithm with a generic base can be written in terms of either the natural base or the common base logarithms using the change-of-base formula:

lnxx log log x  (See Properties of Logarithms.) a lnaa log

Logarithmic Functions

By definition, the logarithmic function yx loga   is the inverse function of the exponential function ya x . It is defined implicitly by the equation xa y . As such, its domain is the set of positive real numbers and its range is the set of all real numbers. Moreover, we have the following identities:

loga x x ax , for x  0 ; loga ax  , for any real number x.

The graphs of logarithmic functions are either growths ( a 1) or decays ( 01a ).

Graph showing logarithmic decay Graph showing logarithmic growth

Exercise Find the domain of f( x ) 1  ln x  2 . Graph this function, find its inverse and any x- or y-intercepts.

Properties of Logarithms

For any base 푎, where 푎 > 0, 푎 ≠ 1, any positive real numbers 푀, 푁, and any real number 푟, the following are always true:

1  푙표푔 1 = 0, 푙표푔 푎 = 1, 푙표푔 ( ) = −1 푎 푎 푎 푎

 푙표푔푎(푀푁) = 푙표푔푎푀 + 푙표푔푎푁

*** This is NOT a formula for 푙표푔푎(푀 + 푁) ***

푀  푙표푔 ( ) = 푙표푔 푀 − 푙표푔 푁 푎 푁 푎 푎

푟  푙표푔푎푀 = 푟푙표푔푎푀

푙표푔푏푀  푙표푔푎푀 = 푙표푔푏푎

[This is the change-of-base formula, valid for any other base 푏 such that 푏 > 0, 푏 ≠ 1. Note that in practice this formula is used with 푏 = 푒, the natural base, or 푏 = 10, the common base, since logarithms with those two bases can be computed on a calculator.]

These properties are used extensively in exponential & logarithmic models to solve equations. Exercises

2x a) Solve the equations 2푙표푔7푥 = 푙표푔79 and e  5 .

b) Show that 푙표푔푎(√푥 + √푥 − 1) + 푙표푔푎(√푥 − √푥 − 1) = 0 for 푥 > 1. c) Approximate the number 푑 that satisfies the equation 3푑 = 4 to the nearest hundredth (this is the fractal dimension of the famous Koch curve). d) Solve the equation 푙표푔5(푥 + 6) = 1 − 푙표푔5(푥 + 2) e) Solve the equation 53xx2 3 2 . f) Solve the equation 4푥 − 2푥 − 12 = 0. [Challenging]

Answers

22 a) 2log7x log 7 9  log 7 x  log 7 9  x  9  x  3 . Note here that x 3 is NOT a solution since it’s not a positive number.

ln5 e22xx5  ln e  ln5  2 x  ln5  x   0.805 2 b) For 푥 > 1,

푙표푔푎(√푥 + √푥 − 1) + 푙표푔푎(√푥 − √푥 − 1) = 푙표푔푎[(√푥 + √푥 − 1)(√푥 − √푥 − 1)] = 푙표푔푎(푥 − (푥 − 1)) = 푙표푔푎1 = 0

ln 4 c) 3푑 = 4 ⇒ 푙푛(3푑) = 푙푛4 ⇒ 푑 ∙ 푙푛3 = 푙푛4 ⇒ d 1.26 ln3

[Note that this solution can be obtained by substituting the natural base logarithms with common base logarithms.]

d) log5x 6  1  log 5 x  2  log 5 x  6  log 5  x  2  1

log5 xx  6  2  1 2 log55xx  8  12  log 5 xx2 8  12  5 xx2 8  7  0 xx 7  1  0  x 1 , since -7 is rejected as an invalid solution

because of the domain of log5 x  6 . e) 5x2 3 3 x  2  ln5 x  2  ln3 3 x  2 xx  2ln5  3  2ln3 x ln5  2ln5  3 x  ln3  2ln3  x ln5  3ln3  2ln3  2ln5 2 ln3 ln5  x  ( 3.212 ) ln5 3ln3

[Check this! Note that this solution can be obtained by substituting the natural base logarithms with the common base logarithms.] f) If we let 2x , then the equation becomes 2   12  0 since x 2 4xx 222  2    . As a result, we have  4   3  0     3,4 . Since is always positive, this implies that  4  2x  4  22  x  2 .