aⁿ = a * a * a * ... * a             // a is repeated n times

a^b * a^c = a^b+c

a^b ¸ a^c = a^b-c

a^c * b^c = (ab)^c

a^-b = 1 / a^b

(a^b )^c = a^bc

Logarithms may be to any base > 0 and  ¹ 1. The most commonly encountered are: common logarithms (base 10, abbreviated “log”); natural logarithms (base e »  2.71828, abbreviated “ln”); and binary logarithms (base 2, abbreviated “lg”).

The following properties hold for logarithms to any base b.  In the following discussion, a and n are real numbers, b a real number > 0 and  ¹ 1.

Let a = bⁿ. Then n = log_b a. Example:
100 = 10² ; 2 = log ₁₀ 100.
12.1824 »  e ^2.5; ln 12.1824 = 2.5.

The following properties follow from the laws of exponents and apply for all logarithms, regardless of base. Let n and m denote positive nonzero real numbers, b as defined above.

 log_b (nm) = log_b n + log_b m

log_b (n / m) = log_b n - log_b m

log_b ( n ^m) =  m log_b n

log_b ( 1/n) =  - log_b n

 Note that these rules hold in the case of fractional exponents as well, i.e. root finding.

log_b (a^1/2) = (log_b a) * 1/2

Converting between bases:
Let log_a x denote a logarithm of x in any valid base a. Let log_b x denote a logarithm of x in some other valid base b.  Further suppose we know the value of log_a b. Then
                        log_a x
 log_b x = ————————
                        log_a b

E.g.,  log x = ln x / ln 10 = lg x / lg 10
 lg y = log y / log 2 = ln y / ln 2

Regardless of base, log_b b = 1; log_b 1 = 0; log_b x is only defined for x > 0.

Examples: Given that ln 5 = 1.6094 and ln 10 = 2.3026, find ln 50, ln 2, ln 8, log 5. Carry all results to four decimal places.

50 = 5 * 10. Therefore ln 50 = ln (5 * 10) = ln 5 + ln 10 = 1.6094 + 2.3026 = 3.9120.

2 = 10 / 5. Therefore ln 2 = ln (10/5) = ln 10 - ln 5 = 2.3026 - 1.6094 = 0.6932.

8 = 2 ³. Therefore ln 8 = ln( 2 ³) = 3 ln 2 = 3 (0.6932) = 2.0796.

log 5 = log₁₀ 5 = ln 5 / ln 10 = 1.6094 / 2.3026 =  0.69895

Note: If you check the above answers by calculator, you will find that some of the above calculations are off by 1 or 2 in the last digit. This is a side effect of roundoff error, an unavoidable phenomenon when doing numerical calculations to only a certain number of significant figures. The usual approach is to carry out the calculations in at least one more decimal place than you need, then round your answer back. Minimizing and compensating for roundoff (and other) calculation errors is a major subject in itself, well beyond the scope of this course. Consult any book on numerical analysis for more information on this large and fascinating subject.
 

Review Mathematical Notation

SI - CS 191 Home Page