LS 1a Fall 2014
Logarithms (“logs”)
“Logarithms” are a generic type of math function that reverse the exponential equation. A
logarithm is a type of function that calculates what exponent (“Y”) a base value (“B”) must be
raised to in order to produce a number (“X”). It effectively asks the question “To what power
must B be raised, in order to yield X?” The answer is the exponent “Y.”
Since logarithms (or “logs,” as the cool kids call them) are effectively reversals of the
exponential function, let’s look at some exponential functions first before flipping them abouts.
Using the same X, Y, and B symbology as in the opening paragraph, an exponential equation
takes on the form X = 𝐵𝑌 . X is the resulting number after a base value has been raised by a
certain exponent.
For example:
(1) If B = 2, and Y = 3, then X = 23 = 8.
(2) If B = 10, and Y = 3, then X = 103 = 1,000.
(3) If B = 10, and Y = -3, then X = 10-3 = 0.001.
Logarithms rearrange these symbols to more directly solve for Y instead of X, using the generic
formula:
Y = log B (X)
We can now rearrange our initial three example, such that:
(1a) If B = 2, and Y = 3, then 3 = log 2 (8).
(2a) If B = 10, and Y = 3, then 3 = log10 (1,000).
(3a) If B = 10, and Y = -3, then -3 = log10(0.001).
Whereas exponential equations most directly ask “what is the result (X) of raising B to the Y?”
logarithmic function ask “what is the power Y that when B is raised to it, produces X?” By
rearranging (or “reversing”) the exponential equation, logs allow for directly solving for the
exponent Y.
In example (1a), the logarithmic equation states that the base value 2 is raised to the third
power, producing the number 8. In the second example, the base value 10 is raised to the third
power, producing the number 1,000. In the third example, the base value 10 is raised to the -3
power, producing the number 0.001.
The logarithm, or “log,” is typically used to refer to the exponent (the “Y” value). So the log of 8
to base 2 is 3; the log of 1,000 to base 10 is also 3; and the log of 0.001 to base 10 is -3.
Unless otherwise specified, it is assumed that logs use a base 10 (these logs are appropriately
named “base-10 logs”). So one can say that the “log of 1,000 is 3.” When logs with a base 10
are assumed, the “B” value in the log relationship is typically not shown:
LS 1a Fall 2014
Y = log10 (B) = log(B)
Logarithms reduce wide-ranging scales to smaller, more manageable scales. Referring to our
second and third examples, the log of 1000 is 3, and the log of 0.001 is -3. Instead of having to
graph or show a wide-ranging effect that could span units from 0.001 to 1,000, a log scale
compresses the range such that the effect could now be show on a scale of -3 to 3 instead of
0.001 to 1,000.
One manner in which we come across logs in LS1a is in the measurement of acidity, in which we
use “pH” as a logarithmic measure for the acidity of an aqueous solution.
There are a few interesting values that we care most about, and it can be helpful to look at a
graph of y = log (x) below:
2
1
0 = log (1)
1
2
3
4
5
6
7
x
1
undefined = log (0)
2
Notice that the log where when x=0 (i.e., “log(0)”) is undefined or “negative infinity,” which is
to say that there is no exponent that could be used to raise a base value to the number zero.
Therefore the bottom of the graph continues to get closer and closer to 0 without ever reaching
it.
Also notice that log (1) equals zero, as the exponent that raises a base value to the number 1 is
zero.
The fact that log(1) = 0 allows us to solve the Henderson–Hasselbalch equation for instances in
which [HA] = [A-], such that [HA]/[A-] = 1:
pKa = pH + log ([HA]/[A-])
LS 1a Fall 2014
pKa = pH + log (1)
pKa = pH (when the concentrations of protonated and deprotonated functional groups are
equal).
This equivalence of pKa and pH when a functional group’s protonated (“HA”) form and
deprotonated (“A-”) form have the same concentration allows us to determine the pKa of
functional groups at the half-equivalence point during titration experiments.
Other instances in which logs come up in LS1a to describe acidity are:
pKa = -log(Ka)
pH = -log([H+]), or in water = -log([H3O+])
One technique that can often be used to solve for log equations is to return it to an exponential
equation by raising both sides by base 10:
Y = log (B)
10Y = B
So that if you know that the pH value of a solution is 2, then
pH = -log[H3O+]
2 = -log[H3O+]
-2
10 = [H3O+]
[H3O+] = 0.01 M = 10 mM
Note that since the logs for both pKa and pH are negative, the exponents you raise 10 to must
be negative.