## Kiss those Math Headaches GOODBYE! ### The Log Blog

It’s that time of year again when Algebra 2 students are all studying logs — not the lumberjack things, but ‘logarithms’ — so I’d like to present a concept that helps students work with logs.

I call it the “Whole-Part-Equality Principle” (as I’ve never seen it named by anyone else … is there a name for it? Anyone know?) But I prefer to call it the “Peter-Paul-Pennies-in-the-Pocket Principle.”

Here’s how it works. There’s Peter, and there’s Paul. We are told that Peter and Paul have no money except pennies, and they transport their pennies only in their right and left pants pockets (if anyone can think of a way to pack this story with even more p’s, please let me know).

Anyhow, we know three additional facts:

1)  The number of pennies that Peter is transporting equals the number of pennies that Paul is transporting.

2) Peter and Paul each have three pennies in their right pants pockets.

3)  Peter and Paul transport their pennies NOWHERE but in their pants pockets.

QUESTION:  What can we conclude about the number of pennies that Peter and Paul have in their left pants pockets?

ANSWER:  It’s obvious, right? While we don’t know how many pennies Peter and Paul could be transporting in their left pants pockets (it could be any number, right?), it is nevertheless clear that they must have the same number of pennies in their left pants pockets.

WHY?  View it like this … If the wholes are the same (the total number of pennies that Peter and Paul each has), and if one of two key parts are the same (the number of pennies that Peter and Paul have in their right pants pockets), then the other parts must also be equal (the number of pennies they have in their left pants pockets).

Why am I bringing this up? To point out an important principle.

This same principle — if the wholes are equal, and if one of their two parts are equal, then the other parts must also be equal — can be used to solve many log and exponent problems.

EXAMPLE 1:  Suppose you have this equation:  log x = log 7.2. What can we conclude? Well, the wholes are equal (meaning the left and right sides of this equation are equal), and the bases of the logs are equal (logs are always base 10 unless another base is given), therefore the remaining parts, the ‘arguments,’ also must be equal. The ‘argument’ is the term after the word ‘log,’ so for this equation the arguments are x and 7.2, and they must be equal … meaning that  x = 7.2.

EXAMPLE 2:   Suppose you have the equation:  log 2^x = log 16. Again, the wholes are equal, and the logs have the same base, so the arguments must be equal. That means that 2^x = 16. Since 2^4 = 16, x = 4, and that’s the answer.

EXAMPLE 3:  Suppose you have the equation:   a^log x = a^log 12.9. Since the wholes (the left and right sides of the equation) are equal, and since the bases are equal as they are both ‘a,’ therefore the only remaining parts, the exponents, must also be equal. So this means that log x = log 12.9. Following the same logic as we used in Examples 1 and 2, this means that x = 12.9.

Any questions? If so, please post as a comment. If not, please use this principle, and enjoy its profound practicality. (OK, I’m done.)

Josh Rappaport is the author of the Algebra Survival Guide and Workbook, which together comprise an award-winning program that makes algebra do-able! Josh also is the author of PreAlgebra Blastoff!, an engaging, hands-on approach to working with integers. All of Josh’s books, published by Singing Turtle Press, are available on Amazon.com