Almost everyone is familiar with the mathematical concept of a mean, even if they know it by its more common name, the average. By summing the terms in a series and dividing the resulting number, you can obtain the mean of a given group of numbers. A logarithmic mean is very much like this. Often used when calculating temperature differences, a logarithmic mean is obtained in much the same way as a simple average, though it does employ a slightly higher level of mathematics associated with logarithms.

The logarithmic mean can only be calculated use two non-negative, real numbers.

Place the two numbers that you will be deriving the mean from in a series by writing them down in sequential order. For example, use 190 and 280, written in that order.

Calculate the value of the natural logarithms (ln) of the numbers using a calculator or slide rule. Write these numbers down. In the example, ln(190) = 5.25 and ln(280) = 5.63.

Calculate the difference of the two numbers you are deriving the mean from by subtracting one, called x, from the other, called y. Calculating the mean of more than two logarithms will require a different formula and higher mathematics, so only use this method for obtaining the mean of two logarithms. Following the example above, 280 - 190 = 90.

Subtract one logarithmic value, called ln x, from the second, called ln y. Use either the log function on your calculator, which can perform the subtraction process in one step, or calculate the value of log x and log y individually and subtract these two numbers from one another. Keep track of the order in which you are subtracting the numbers. Continuing with the example, 5.63 - 5.25 = 0.38

Divide the difference of x and y by the difference of ln x and ln y. Make sure that x and y are in the same order in the quotient and denominator of the fraction. In the example problem, 90/0.38 = 236.84. The logarithmic mean is 236.84.

#### Warnings

#### References

- Wolverine Tube Heat Transfer Data Book: The Mean Temperature Difference
- "Fluid Mechanics and Transfer Process"; Kay, et al.; 1985

#### Photo Credits

- Brand X Pictures/Brand X Pictures/Getty Images