Logarithms, Exponential Functions, and Graphing – A Primer ...

1B Logs and Graphing Primer.doc

Logarithms, Exponential Functions, and Graphing – A Primer Discussion A few topics in Chemistry require the use of logarithms. This exercise is meant to be a review. For those students without prior knowledge in this area, there is no reason to be intimidated by logarithms. Most simply, logarithms are mathematical functions that extract the exponent from the exponential representation of a number. Antilogarithms are literally functions that “undo” the taking of a logarithm. There are two common forms of logarithms: logarithms to the base 10 (log10 or simply log) that are called common logarithms and logarithms to the base e (loge or simply ln) that are called natural logarithms. We will discuss common logarithms initially because they are easier to explain and understand and then we will extend the explanation to natural logarithms.

Common Logarithms Suppose we have a number N = 10X. Then log10N = X, the log operator simply “extracted” the exponent X from the exponential representation of the number. From this, it is easy to see that the antilogarithm of X is 10X = N. Taking logarithms of numbers which can be represented by integer powers of 10 is particularly easy, for example, 2

log 100 = log 10 = 2 –3 log 0.001 = log 10 = – 3 0 log 1 = log 10 = 0 Evaluating antilogarithms of integers is even easier. 5

antilog 5 = 10 = 100000 –2 antilog –2 = 10 = 0.01

0

antilog 0 = 10 = 1 Note that although you cannot take the logarithm of a negative number or zero, you can take the antilogarithm of any real number. Taking the logarithm of a number that cannot be wholly represented by an integer power of 10 is less obvious. For example, 0.3010 you would have to know that 10 = 2.000 in order to find log 2.000 = 0.3010. In “the old days”, it was necessary to use log tables to find the logarithms of such numbers. Fortunately, now all scientific calculators have “log” or “log x” or something similar to find the logarithm of any positive number within the computational range of the calculator. You should be able to find log 0.05 on your calculator (it’s – 1.3) and log 3000 is 3.4771. Finding the antilogarithm is just as easy. The antilog function on most calculators is activated by pressing the “2nd” function key, the “INV” function key, or the “shift” function key followed by the “log” key. 10X is printed above the log key on the face of most calculators. You should be able to find the following antilogs by using your calculator: 2.6020

antilog 2.6020 = 10 = 399.9 4.3010 4 antilog 4.3010 = 10 = 2.000x10

antilog –1.6990 = 10

–1.6990

= 0.02000

Natural Logarithms Use of natural logarithms (ln) is analogous to the use of common logarithms, except that the base of natural logs is the irrational number e (e = 2.71828...). This is a number like pi that arises naturally in mathematics and can’t be represented by the ratio of any two integers (hence irrational). All scientific calculators have a natural log key usually labeled “ln” or “ln x”, and an anti-natural log key which is usually labeled eX or exp() and is accessed in the same way as the 10X. Taking natural logs and antilogs is the same as taking common logs and antilogs except that one uses the “ln” and “eX” instead of the “log” and “10X” keys, respectively.

1B Logs and Graphing Primer.doc

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Last updated 26-Jan-08

1B Logs and Graphing Primer.doc

You should verify that you know how to find natural logarithms with the following: ln 35 = 3.56

ln 10 = 2.30

ln 8.64x10–2 = – 2.449

ln 623.7 = 6.4357

Also check that you can get the following results on your calculator: 1.36

antiln 1.36 = e = 3.9 = exp(1.36) 0 antiln 0 = e = 1 = exp(0)

antiln –0.362 = e–0.362 = 0.696 = exp(-0.362)

In the natural sciences, many derivations result in relationships involving natural logarithms. For many years, before scientific calculators were widely available and inexpensive, common logarithms were widely used in preference to natural logarithms. The natural log of a number, a, is 2.303 times the common log. ln(a) = 2.303 log(a)

Mathematical Operations Using Logarithms and Exponential Functions Sometimes it is necessary or desirable to manipulate logarithms. Some relationships involving logarithms are as follows: 10a x 10b = 10(a+b) log(10a) = a 10log(a) = a log(ab) = log(a) + log (b) log(1/a) = – log(a) log(a/b) = log(a) – log(b) n log(a ) = n log(a) log(a1/n) = (1/n) log(a)

ea x eb = e(a+b) ln(ea) = a eln(a) = a ln(ab) = ln(a) + ln(b) ln(1/a) = – ln(a) ln(a/b) = ln(a) – ln(b) n ln(a ) = n ln(a) ln(a1/n) = (1/n) ln(a)

These relationships apply to logarithms of all bases.

Significant Figures in Logarithms and Antilogarithms The rules for significant figures in logarithms are not as direct as one would like: 1.

When taking a logarithm, retain as many digits to the right of the decimal place in the result (when written in decimal form) as there are significant digits in the original number. log(34508) (5 sig. digits) = 4.53792 (5 decimal places) –6 ln(1.45x10 ) (3 sig. digits) = –13.444 (3 decimal places)

2.

When taking an antilogarithm of a number (a) if the number (when written in decimal form) has digits to the right of the decimal place, retain as many significant digits in the result as there are digits to the right of the decimal place in the original number. antilog(1.678) (3 decimal places) = 47.6 (3 sig. digits) –44 exp(–100.90) (2 decimal places) = 1.5x10 (2 sig. digits) (b) if the number (when written in decimal form) has no digits to the right of the decimal place, retain only 1 significant digit. (Technically, in this case, the result has no significant digits! What this means is that the uncertainty in the reported answer is relatively large.) 56

antilog(56) = 1x10 –45 exp(-103) = 2x10

A note of caution when doing calculations involving logarithms and antilogarithms: Carry extra digits through each step of a calculation and round off the final result to the correct number of significant digits. Rounding off at each step can cause a significant rounding error in the end since logs and exponential functions can be very sensitive to small errors.

1B Logs and Graphing Primer.doc

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Last updated 26-Jan-08

1B Logs and Graphing Primer.doc

Exponential Functions

The functions ex = exp(x), e-x = exp(-x), e1/x = exp(1/x), and e–1/x = exp(-1/x) are very common in chemistry. They are especially useful in kinetics and thermodynamics. It is a good idea to recognize the shape of these functions and their limits for small and large values of x.

Exponential Functions and Graphing Many of the natural behaviors we observe in chemistry have mathematical expressions that are exponential (e –x) in nature. This primer is intended to help you gain a better understanding of exponential functions and how these functions can be “linearized” to form straight-line graphs. The linearization of exponential functions is often done in introductory textbooks since students are more familiar with linear equations. To show the relationship between the nonlinear exponential decay function (e–x) and its straight-line counter part (ln(y) = mx+b) we will examine the decay of a radioactive isotope over time. The relationship between the remaining mass (not yet decayed) and time for a radioactive isotope follows an exponential decay and has the form m(t) = m0e-kt

(1)

where m is the remaining mass, t is time, m0 is the mass at time zero, and k is the radioactive decay constant. A graph of the decay of Iodine-131, 131I, is shown. Using Graphical Analysis the data points are fit with the nonlinear function, m(t) = m0e-kt , the fitting constants, m0 and k are found from the “best-fit” to the data points, see graph. The mass at time zero, m0, is predicted to be 9.5 mg and the radioactive decay constant, k, is found to be 0.067/hr. Therefore the equation of the decay is m(t) = (9.5 mg)e–0.067t

Linear form of Exponential Functions A linear form for the decay of a radioactive isotope can be derived from the exponential decay equation (1). If the natural log of both sides is taken we obtain ln(m(t)) = -kt + ln(m0) (2) Equation 2 is in the form a straight line with y-axis = ln(m(t)), x-axis = time, slope = -k and y-intercept of ln(m0). The linearized data and straight-line graph is shown. The linear equation of the decay is ln(m(t)) = -0.069t + 2.26 1B Logs and Graphing Primer.doc

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1B Logs and Graphing Primer.doc

The slope and y-intercept determine the radioactive decay constant and the initial mass at time zero, respectively. The decay constant, k, is equal to -slope = 0.069/hr. The mass at time zero is ey-int = 9.6 mg. Notice the values for k and m0 from the linear fit are slightly different than those obtained from the nonlinear fit. The reason for this slight difference is not important at this time and can be neglected.

Exponential and Logarithmic Functions – UNITS! Mathematically, when we take the logarithm or exponential of a number, the number itself must be unit less, since units cannot be carried through logarithm and exponential functions. When the exponential term was –kt in the previous example, the units of k and t must cancel each other. Since t has units of time, k must have units of 1/time, in our case 1/hr. The same is true for logarithms. If we take the logarithm of a mass (mg) the resultant value is unit less, the mass unit is dropped. In the last figure the y-axis is unit less since the numbers represent logarithmic values. However, the original units of mass of mg are clearly noted on the y-axis (mass in mg). The units of the slope (y/x) then become 1/time as they should since the slope has the same units as the decay constant, k.

1B Logs and Graphing Primer.doc

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