**ME33S Advanced uses of logarithmic functions**

Log and antilog functions

Before calculators like the ME33S became easily available, logarithms were commonly used to simply multiplication. They are still used in many subjects, to represent large numbers, as the results of integration, and even in number theory.

The ME33S has four functions for calculations with logarithms.

These are the “common” logarithm of “x”, $, its inverse, !, the “natural” logarithm of “x”, & and its inverse, #.

Common logarithms are also called “log to base 10” and the common logarithm of a number “x” is written

LOG10 x or just LOG x

Natural logarithms are also called “log to base e” and the natural logarithm of a number “x” is written

LOGe x or LN x

Logarithms can be calculated to other bases, for example the log to base two of x is written

LOG2 x

Some problems need the logarithm of a number to a base n, other than 10 or e. On the ME33S these can be calculated using one of the formulae

LOGn x = LOG10 x ÷ LOG10 n

LNn x = LNe x ÷ LNe n

! and # are also called “antilogarithms” or “antilogs”. # is also called the “exponential” function or “exp”. Apart from being the inverses of the log functions, they have their own uses. ! is very useful for entering powers of 10, especially in programs where the } key can not be used to enter a power that has been calculated. # is used in calculations where exponential growth is involved. 1# is a quick way to type the value of e.

The ) function can be seen as the base “n” antilog function. If 10x is the inverse of log10 x and ex is the inverse of loge x, then yx is the inverse of logy x.

Practice using log and antilog functions

Example 1: Find the common logarithm of 2.

Solution: In RPN mode type 2¹$

Figure 1

In algebraic mode type 2¹$Ï MEcalculators - 2 - ME33S Advanced uses of logarithmic functions - Version 1.0

MEcalculators

ME33S Advanced uses of logarithmic functions

Figure 2

Answer: The common logarithm of 2 is very nearly 0.3010.

Example 2: A rare species of tree has a trunk whose cross-section changes as 1/x with the height x. (Obviously this breaks down at ground level and at the tree top.) The cross section for any such tree is given by A/x, where A is the cross-section calculated at 1 meter above the ground. What is the volume of the trunk between 1 meter and 2 meters above ground?

Solution: The volume is obtained by integrating the cross-section along the length, so it is given by the integral:

Figure 3

It is possible to evaluate this integral using the ME33S integration function, but it is much quicker to note that the indefinite integral of 1/x is LN x. The result is therefore

V = A (LN2 – LN1)

Since LN 1 is 0, this simplifies to

V=A LN2

In RPN or algebraic mode type 2&. In algebraic mode follow this with Ï

No one is likely to measure tree heights to an accuracy of more than three significant digits, so set the

ME33S to display the answer with just 3 digits after the decimal point, by pressing Þ13

Figure 4

Answer: Figure 4 shows that the log to base e of 2 is close to 0.693, so the volume is 0.693A cubic meters.

Example 3: What is the log to base 3 of 5? Confirm the result using the ) function.

Solution: Using the equations given above, the log to base 3 of 5 can be calculated as (log10 5)/(log10 3).

In RPN mode, press: 5¹$3¹$¯

In algebraic mode, press: 5¹$¯3¹$Ï

MEcalculators - 3 - ME33S Advanced uses of logarithmic functions - Version 1.0

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*12* 33S Advanced uses of logarithmic functions

Figure 5

That this is correct can be confirmed if the following keys are pressed.

In RPN mode: 3w)

In algebraic mode: 3)¹ÍÏ

Figure 6

Answer: The log to base 3 of 5 is 1.465 within the current accuracy setting of the calculator, as shown by Figure 5. Calculating 3 to this power gives 5.000 which confirms that the correct value for the log had been obtained.

Example 4: An activity of 200 is measured for a standard of Cr51 (with a half-life of 667.20 hours). How much time

will have passed when the activity measured in the sample is 170? The formula for half-life computations

is shown in Figure 7.

Figure 7

Solution: Rearrange the equation to solve for t, as in Figure 8.

Figure 8

Now calculate t. In RPN mode:

667Ë2Ï170Ï200¯&¸Ë5&¯

In algebraic mode:

667Ë2¸ºy170¯200º|&¯Ë5&Ï

Figure 9

Answer: 156.4352 hours. Figure 9 shows the result in RPN mode.

Note: When you have finished this example, press Þ4 to return the display to Standard display mode.

**Note: Please ignore the above information, in case it has been informed or circulated.**

*Edited: 10 Nov 2009, 2:53 p.m. after one or more responses were posted*