This help file deals with significant figures when using
logarithims and manipulating logarithms when numbers are
too large for your calculator to display.


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Significant figures dealing with logarithms can be tricky.

When taking the base 10 log (also known as "common" logarithm)
of a number the sig. fig. info winds up to the right of the
decimal place.  For example, if your number is

4.8 x 10^2 (2 s.f.)

then

log(4.8 x 10^2) is 2.6812.


This should be reported as 2.68.  The "2" to the left of the decimal comes
from the power of 10 (which has no s.f. info).  The "68" to the right of
the decimal signifies the number of s.f. in the original number.  To see
this lets break the number above into two parts using the rules of logs.

log(4.8 x 10^2) = log(4.8) + log(10^2)

                = 0.6812 + 2

		= 2.6812

		= 2.68


In other words, when taking the log of a number your s.f. info winds up to
the RIGHT of the decimal and is given by the number of decimal places shown.
If then original number has 4 s.f. the log of the number should have 4
decimal places.  Thus, what ever number of s.f. you have in the original
number, when you take the log of the number you should show the same number
of decimal places in the log as there were s.f. in the original number.

Appendix A in the textbook also covers this.  

This is easily shown for the base 10 logarithm.  How about natural logarithms
(base "e"), "ln"?  The ln and log of a number are related through the ln(10),

	ln(a) = 2.303 log(a)

The "2.303" comes from ln(10).  We usually truncate it as 2.303 but we could
write it as,

	ln(a) = ln(10) * log(a)

Doing it this way ln(10) has many more digits than the four (2.303) usually
shown in the conversion between the two logarithems.  Done this way it would
not effect our sig. fig. in any way for most of what we do.  Thus, we can use
the same rule for sig. fig. for natural log, ln, as we do for base 10 log.

Remember when taking the log of a number the s.f. info winds up to the
RIGHT of the decimal point.  So log(x) and ln(x) should have the same number
of DECIMAL places as there are s.f. in the original number.


What about going the other way, i.e. taking the antilog?

Lets say we have something like,

log(X) = 1.047

and we are trying to find "X".  We simply take the antilog (the inverse
function of log).

  X = 10^(1.047) = 1.114294534 x 10^1  (displayed in calculator)

How many s.f. should "X" have?

Following the rules above in reverse, since the 1.047 has 3 decimal places
then "X" should have 3 s.f.,

  X = 1.11 x 10^1   (11.1)

Same if the log is a negative number,

log(X) = -1.047

  X = 10^(-1.047) = 8.974287945 x 10^-2 (displayed in calculator)

Again, as above, since the -1.047 has 3 decimal places "X" has 3 s.f.

  X = 8.97 x 10^-2


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Sometimes the log(X) is a very large number, such as 123.64,

  log(X) = 123.64

and you need to find "X".

Taking the antilog one gets,

  X = 10^123.64

However, if you do this on your calculator (at least most calculators and
particularly the ones allowed for quizzes and exams) you get:

  "Error"

The number is too large for the calculator to display.  What can you do?
Use what you learned many years ago in math/algebra and do some of it the
old-fashioned way, by using pencil and paper.

  X = 10^123.64 = (10^0.64 * 10^123) =  4.365158 x 10^123

How about the s.f. in "X"?

Since the log(X) = 123.64 has 2 decimal places "X" should have 2 s.f.,

  X = 4.4 x 10^123