Logarithms are extremely important in environmental statistics. Chemical concentrations in the environment are often better understood and analyzed in terms of the logs. Estimating average values often requires solving equations that are greatly simplified by using logarithms. Exponentials (the inverses of logarithms) are needed in the mathematical expressions of probability density functions. Many batches of data are so skewed that they are best represented in logarithmic scales. Therefore, you need to know what logarithms are, how to compute with them, and how to understand intuitively the relationship between any number and its logarithm.
The (natural) logarithm of x, written ln(x), is the (signed) area under y = 1/t from t=1 to t=x. For example, let's estimate ln(2) using two trapezoids. The area of a trapezoid is its height times the average of its bases. In the figure, the trapezoids are placed on their sides, so their heights are each 1/2 and their bases are given by the y=1/t formula:
ln(2) = 1/2 * (1/1 + 1/1.5)/2 + 1/2 * (1/1.5 + 1/2)/2 = 17/24 = 0.71 (approximately). This is a little high, as the figure suggests, but evidently it's close.
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Two trapezoids approximate the area under the curve from 1 to 2. |
The laws of logarithms follow readily from this definition. In words, these laws state that logs convert multiplication to addition, from which it follows they convert division to subtraction, and exponentiating (raising to powers) to multiplication. The roles of 0 for addition and 1 for multiplication are identical (any number when added to 0 or multiplied by 1 is unchanged). Therefore, for any numbers a and b where all the relevant expressions are defined,
ln(1) = 0
ln(a*b) = ln(a) + ln(b)
ln(1/a) = -ln(a)
ln(a/b) = ln(a) - ln(b)
ln(ab) = b*ln(a)
It's obvious that ln(1) = 0 because there's no area. The remaining laws follow from this and the multiplication law; namely, that ln(a*b) = ln(a) + ln(b). For example, ln(a) + ln(1/a) = ln(a * 1/a) = ln(1) = 0, so ln(1/a) = -ln(a). Therefore ln(a/b) = ln(a * 1/b) = ln(a) + ln(1/b) = ln(a) - ln(b). You can take the last law as the definition of raising a number to a power, and then check that it agrees with the elementary definition (for positive integral b) that ab is a multiplied by itself b times. The definition (5), however, works for positive values a and any exponent b, positive or negative, integral, fractional, or irrational.
Let's see why the multiplication law is true. Consider the area under y = 1/t from b to a*b (rather than from 1 to something). If we change this shape by "squeezing" the x-axis by a factor of b (that is, divide all x-values by b), then we divide the area by b and the graph becomes y = 1/(b*t). We can make up for the squeeze by expanding the y-axis by the same factor of b (that is, multiply all y-values by b). This multiplies the area by b and the graph becomes y = b * 1/(b*t) = 1/t. That's the same graph we started with. The original area is left unchanged because the y-expansion exactly compensates for the x-shrinkage. Although the curve is unchanged, the x-shrinkage changed the x-limits. They now go from b/b = 1 to a*b/b = a. This area, by definition, is ln(a).
This reasoning shows that we could just as well have defined ln(x) as follows:
Pick any positive number a. Ln(x) is the signed area under y = 1/t from t=a to t=a*x.
Now the fundamental law of logarithms follows immediately by making a vertical cut in the region from 1 to a*b exactly at the value a and comparing areas:
ln(a*b) = area from 1 to a*b = (area from 1 to a) + (area from a to a*b) = ln(a) + ln(b), which is what we wanted to show.
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Geometric demonstration that the log of a product is the sum of the logs. |
Note that since ln(x) is the (signed) area of 1/t from 1 to x, we cannot cross the vertical asymptote of y = 1/t at t=0, so that logarithms of negative numbers are not defined.
With a little more work (using estimates of large logs much as we estimated ln(2) at the beginning) it can be shown that ln(x) becomes arbitrarily large, albeit very slowly, as x grows large. Since ln(1/x) = -ln(x), logarithms become arbitrarily large and negative as x approaches zero. This is useful to know, because it implies
Every number is the logarithm of some (positive) number, so you can always answer the question, "the logarithm of what number is x?".
The answer to this question is the exponential of x, written exp(x). For example, exp(0) = 1, exp(0.7) = 2 (approximately).
You can improve your intuition about logarithms by learning to calculate them. Approximating the area under the curve is difficult and chancy, a task requiring the creativity of a good Renaissance mathematician. There are better ways.
One approach begins by relabeling the t axis. Subtract 1 from each value of t. This shifts the curve y = 1/t to the left by one unit. The equation of the curve is now y = 1/(1+t), because we first have to add that 1 back to each t before computing y.
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The area under the curve y = 1/(1+t) from 0 to x is the area under the curve y = 1/t from 1 to 1+x, which is ln(1+x). |
Algebra gives us
1/(1+t) = 1 - t + t2 - t3 + ... + (-t)N + ...
This infinite series converges absolutely only when the size of t (|t|) is less than 1, for otherwise the successive terms just keep growing in size and obviously cannot converge.
Nevertheless, as you will see, it is very useful to be able to compute ln(1+x) for small values of x. To do so, we need to find the area under y = 1/(1+t) from 0 to x (remember, we shifted all t-values left by 1, so the starting point of 1 becomes 0 and the ending point of 1+x becomes x; see the figure above). For values where the series for 1/(1+t) converges absolutely, we can integrate it term by term. Recalling from elementary calculus that the area (integral) of tN from 0 to x is xN+1/(N+1), we get immediately
ln(1+x) = x - x2/2 + x3/3 - ... -(-x)N+1/(N+1) + ...
This is the Taylor series expansion of ln() about the value x = 1. Remember, it is valid only when |x| is less than one, which means it works only for computing logs between 0 and 2. It works really well when the number for which we want the log is close to 1, for then x is very small and the powers of x rapidly get much smaller.
When x is positive, so we're focusing on computing logarithms of numbers between 1 and 2, the Taylor series alternates in sign: we start with x, subtract something smaller, add something yet smaller, and so on. In this case, the error we make by taking the first few terms of the Taylor series is no larger than the size of the next term in the series. Furthermore, the sign of the next term tells us whether our estimate is low or high: if the next term is positive, that means we have underestimated the value; if the next term is negative, we have overestimated the value.
For example, if we estimate ln(1.2) = ln(1 + 0.2) by the first two terms 0.2 - 0.22/2 = 0.18, the error is no greater than the size of the next term, which is |0.23/3| = 0.00266... . Thus, 0.18 is accurate to two decimal places and we know that it is slightly too low. (The log of 1.2 is 0.18232...).
If you memorize
ln(2) = 0.693... (0.7 is often good enough)
ln(10) = 2.302... (2.3 is often good enough)
then you can readily compute the natural logarithm of any number to a good approximation using the laws of logarithms and the Taylor series expansion.
It is sometimes helpful, but not really necessary, to know
e = 2.718281828...
where e is defined by ln(e) = 1.
Another approximation worth remembering is ln(20) = ln(2 * 10) = ln(2) + ln(10) = 2.995... = 3.00.
From the Taylor series
ln(1 + x) = x - x2/2 + x3/3 - x4/4 + ... ,
which is easy to remember--recall the derivation where we passed from 1/(1+t) to 1 - t + t2 + ... to t - t2/2 + t3/3 - ... by integration--we get the useful two-decimal place approximations
ln(1 + x) = x for |x| < 0.1 (this is an overestimate when x > 0, because the next term would be negative)
ln(1 + x) = x - x2/2 for |x| < 0.23 (this is an underestimate when x > 0, because the next term would be positive)
Here are some examples.
ln(3) = ln(2 * 5/4 * 6/5) = ln(2) + ln(1 + 0.25) + ln(1 + 0.20) = (underestimate) 0.693 + 0.25 - 0.252/2 + ... + 0.20 - 0.202/2 + ... = 0.69 + 0.25 - 0.031 + 0.20 - 0.02 = 1.092 (the correct answer is 1.0986...). This example also illustrates why, if you want your answer to two decimal places, you really should compute intermediate results to three decimal places.
Here are some more examples of computing logs to two decimal places.
ln(4) = 2*ln(2) = 1.39.
ln(5) = ln(10/2) = ln(10) - ln(2) = 2.302 - 0.693 = 1.61.
ln(1/2) = -ln(2) = -0.69.
ln(100) = ln(102) = 2*ln(10) = 2*2.302 = 4.60.
ln(1000) = ln(103) = 3*ln(10) = 6.91.
ln(1 000 000) = 6*ln(10) = 13.82.
ln(0.8) = ln(23 / 10) = 3*ln(2) - ln(10) = 3*0.693 - 2.302 = -0.22 (rounded).
It's very easy to compute the logarithms of nastier numbers, too: just look for the closest number for which you can reliably estimate the log from values you already know, then use the Taylor series approximation:
ln(0.79) = ln(0.8 * (1 - 1/80)) = -0.22 - 1/80 = -0.23.
ln(4.4) = ln(4 * 1.1) = ln(4) + ln(1 + 0.10) = 1.386 + 0.10 - 0.01/2 + ...= 1.48.
ln(17) = ln(16 * (1 + 1/16)) = 4*ln(2) + 1/16 - ... (because 1/16 is much less than 0.1, we need only one term in the series) = 2.772 + 0.0613 = 2.83.
ln(Pi) = ln(3 + 1/7, approximately) = ln(3 * (1 + 1/21)) = 1.10 + 1/21 - ... = 1.15 (a slight overestimate; the correct value is 1.14 to two decimal places).
If you're not being finicky and just want to know the logarithm roughly, life is easy:
ln(93) = ln(100 * 0.93) = 2 * ln(10) + ln(1 - 0.07) = 4.6 - 0.07 = 4.5, roughly.
ln(560 000) = ln(5.6 * 105) = ln(5 * 1.12 * 105) = ln(106 / 2 * 1.12) = 6 * 2.3 - 0.7 + 0.12 = 13.2 (it's actually 13.236...).
You should begin to see why you need only four simple things for rough computation of logs:
The computation is so straightforward you can readily accomplish it with mental arithmetic if you just want the log to one decimal place, which is often good enough.
An earlier version of this material, which is more elementary in nature, is available at Geometric means, exponentials, and logs.
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This page was created 16 February and last updated 3 May 2001.