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Logarithm Formulas - log formula sheet

Logarithm Formulas
Expansion/Contraction Properties of Logarithms
These rules are used to write a single complicated logarithm as several simpler logarithms (called "ex-
panding") or several simple logarithms as a single complicated logarithm (called "contracting"). Notice that
these rules work for any base.
loga(xy) = loga(x) + loga(y) (multiplication inside can be turned into addition outside, and vice versa.)
x = loga(x) - loga(y) (division inside can be turned into subtraction outside, and vice versa)
loga y
loga(xn) = n ? loga(x) (an exponent on everything inside can be moved out front, and vice versa)
Change of Base Formula
This formula is used to change a less helpful base to a more helpful one (generally base 10 or base e, since
these appear on your calculator, but you can change to any base). In the formula below, a is the current
base of your logarithm, and b is the base you would like to have instead.
loga(x) = logb(x)
logb(a)
Cancellation Properties of Logarithms
These rules are used to solve for x when x is an exponent or is trapped inside a logarithm. Notice that
these rules work for any base.
loga(ax) = x (this allows you to solve for x whenever it is in the exponent)
aloga(x) = x (this allows you to solve for x whenever it is inside a logarithm)
Logarithm Problems
1. Expand each expression (use expansion properties to expand as much as possible).
a. log3(xy)
Solution:
log3(xy) = log3(x) + log3(y) (multiplication rule)
x
b. log7 y
Solution:
x = log7(x) - log7(y) (division rule)
log7 y
c. ln(x5)
Solution:
ln(x5) = 5 ln(x) (exponent rule)
d. log x2y3
z4
Solution: Whenever multiple rules seem to apply, you should always do them in reverse order
of operations. Imagine you knew that x = 5, y = 4, and z = 2, and you wanted to
evalaute x2y3
z4 . Your work would be
52 ? 43 25 ? 64 (take care of exponents)
24 = 16
= 1600 (multiply numbers in numerator)
16
= 100 (divide)
When expanding logarithms, you'll want to work in reverse. In this example, that
means apply division rule, then the multiplication rule, then the exponent rule.
log x2y3
z4 = log(x2y3) - log(z4) (division rule)
= (log(x2) + log(y3)) - log(z4) (multiplication rule)
= log(x2) + log(y3) - log(z4) (parentheses don't matter here)
= 2 log(x) + 3 log(y) - 4 log(z) (exponent rule)
e. ln x2-y
wz2
Solution:
ln x2 - y
wz2 = ln(x2 - y) - ln(wz2) (division rule)
= ln(x2 - y) - (ln(w) + ln(z2)) (multiplication rule)
= ln(x2 - y) - ln(w) - ln(z2) (distribute the minus sign)
= ln(x2 - y) - ln(w) - 2 ln(z) (exponent rule)
This problem has a couple tricky parts.
One is that you need to be careful about parentheses when you apply rules. When
you have "- ln(wz2)", you're subtracting the entire quantity, so when you expand
ln(wz2) to ln(w) + ln(z2), you want to be subtracting this entire quantity. This is
accomplished by writing "-(ln(w) + ln(z2))" rather than "- ln(w) + ln(z2)" (take
a moment to convince yourself that the parentheses really make a difference here).
Until you become comfortable with these kinds of problems, it's better to always write
down parentheses and then decide later that you didn't actually need them. Having
they are needed does.
The other thing to realize is that there is no rule to expand ln(x2 - y). It is tempting
to think that the division rule applies here, but read the rule very closely. It involves
"log minus log", not "subtraction within a log". The same is true of addition; there
is no rule to handle "addition within a log". Since the term cannot be expanded, we
simply leave it as it is.
2. Contract each expression (use contraction properties to write as compactly as possible).
a. log2(x) + log2(y)
Solution:
log2(x) + log2(y) = log2(xy) (multiplication rule)
b. log9(x) - log9(y)
Solution:
x (division rule)
log9(x) - log9(y) = log9 y
c. 5 log(x)
Solution:
5 log(x) = log(x5) (exponent rule)
d. ln(x) + ln(y) - 2 ln(z)
Solution: When we expanded logarithms, we worked in reverse order of operations. Since
contracting is the opposite of expanding, we should go back to the normal order of
operations. For this example, that means taking care of the "times 2" in front of the
"ln z", then working from left to right after that.
ln(x) + ln(y) - 2 ln(z) = ln(x) + ln(y) - ln(z2) (exponent rule)
= ln(xy) - ln(z2) (multiplication rule)
= ln xy
z2 (division rule)
e. 2 log(2x) - 3(log(y) + log(z))
Solution:
2 log(2x) - 3(log(y) + log(z)) = 2 log(2x) - 3 log(yz) (multiplication rule)
= log((2x)2) - log((yz)3) (exponent rule)
= log(4x2) - log(y3z3) ("distribute" the exponent, if desired)
4x2
= log y3z3 (division rule)
Just like we saw earlier with the minus sign, you have to be careful when contracting
using the power rule. When we move the number up that is to become an exponent, we
want the entire contents of the logarithm to be affected by this new exponent. That's
why we write, for example, "log((2x)2)" rather than "log(2x2)" (take a moment to
convince yourself that the parentheses really make a difference here).
f. log4(x)+2 log4(y)
3 log(x)-log(y)
Solution:
log4(x) + 2 log4(y) = log4(x) + log4(y2) (exponent rule)
3 log(x) - log(y) log(x3) - log(y)
log4(xy2) (multiplication rule)
= log(x3) - log(y)
= log4(xy2) (division rule)
log x3
y
It is tempting to try to contract the remaining two logarithms, but read the division
rule very closely. It allows us to do something with "division within a log", not "log
divided by log". There is no rule to handle this situation, so we simply leave it as it
is. Similarly, there is no rule to handle "log times log", and so we cannot do anything
to simplify problems where this appears.

What is a log formula?log a = r A natural logarithm is written ln and a natural logarithmic equation is usually written in the form: ln a = r So, when you see log by itself, it means base ten log. When you see ln, it means

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