Proving is a process - property of logarithms
The purpose of this article is to talk about proving. Many kids fear proving and just cannot do it. This may stem from several reasons. Remember that proving a math thing is like solving a problem: it is a process. It many times is a process of trial and error. The proofs that you see in math books are polished and finished and look like they go so smoothly from step to step. It makes you wonder how in the world did they suddenly understand to use this or that as the next step??? A typical math curriculum does not talk much about proving until in high school students are all of a sudden expected to produce such fine and dandy proofs in geometry.
In reality, solving math problems or proving something is kind of similar to many other trades where people solve problems and use their reasoning powers just naturally without any anxiety over it. For example, a doctor has a problem every time a patient comes in with mysterious symptoms. An auto mechanic has to solve the problem as to why the car does not run smoothly. A housewife may be solving the problem of how to make these special tasting cookies she tasted a week ago somewhere. Yet these people don't lose hope or give up when facing problems. They start somewhere, see if it works, and go from there. Trial and error. Another trial.
In times past, kids used to take on their father's trade pretty much. They would observe the father work, and no doubt they would see dad face problems dad didn't know how to solve at first go. So the children would see a problem solving process first hand!
This exact same thing can be very useful in learning math. Try sometimes, take a problem that your child couldn't solve, and "think it aloud" for her. Just make plain to her the whole process of mistakes and several different ideas and maybe checking back to reference books. It can be so eye-opening to youngsters that math problems are usually NOT solved by some mysterious "AHA!" thing where at first sight you know the answer.
And here you will see an example of a problem solving process... And what follows is my true thinking process in solving this problem. It is exactly how I thought through it.
Somebody asked me how to prove that
n log_{a}M = log_{a}M^{n}
I tackled the problem as a math teacher that has been away from logarithms for 5-6 years. I knew that antilogarithms, or raising the number a to some power would probably be helpful since I know that in general, logarithms often get solved by the exponentiation process. So as a try to make things simpler, I decided to call the complex looking thing on the right side x. So I wrote
x = log_{a}M^{n} from which using the definition of logarithm you get a^{x} = M^{n}.
Okay, that looked simpler, so I did the same for the other side of the original equation, calling it y:
y = n log_{a}M from which dividing both sides by n, you get y/n = log_{a}M, and now using the definition of logarithm I got a^{y/n} = M.
So this far using only the definition, I had these two equations a^{x} = M^{n} and a^{y/n} = M. They kind of looked similar. I saw quickly that taking the nth root on the first equation, it would become a^{x}^{/n} = M (because nth root is the same as number to the 1/n th power). And if you have a^{y/n} = M and a^{x}^{/n} = M, then obviously a^{y/n} = a^{x/n}. It looked good, because I do know the fact that in this situation, the exponents must be the same. So y/n = x/n and so y = x. And that's what I was supposed to prove, because I chose x and y to be the two sides in the original problem.
BUT, my proof did certainly not look like one of those textbook "dandy" 2-3 steps proofs. Could I arrange it to look like that? Well, looking back, if I take antilogarithms (or raise a to these powers) of both sides of this supposed equation,
n log_{a}M ? log_{a}M^{n} |
a^{ n logaM } ? a^{logaMn} |
Okay, it looks kind of complicated, BUT it also looks like it could go somewhere, because I can try use the property of exponents that c^{dz} = (c^{d})^{z}.
a^{ n logaM } ? a^{logaMn} |
(a^{ logaM})^{n} ? a^{logaMn} |
Well, the left side is now simply M^{n} (since a^{ logaM} is M) - and the right side simplifies too, because if you think of the M^{n} as a single entity or as "something", then a^{logaMn} is simply M^{n}. (It's a raised to base a logarithm of "something", which since those are opposite operations, it simplifies to "something").
But this still wasn't the textbook polished proof, because if you noticed, I was using ?, or question mark, instead of equal sign, to mark that I don't yet know if the two things are equal. To prove, I would start from something that is known for sure, and proceed from there to whatever I need to prove, whereas above I took what I was supposed to prove and started playing around, hoping to get to a result that is true. I did get to something that is true and so I can rewrite the process in the opposite order and that should look like a finished proof. So starting from the known fact that
M^{n} |
^{=} |
M^{n} |
Then, write M = a log_{a}M and M^{n} = a log_{a}(M^{n}) | |
(a^{ logaM})^{n} | = | a^{loga(Mn)} | Then on left side use the property of exponents that c^{dz} = (c^{d})^{z}. | |
a^{ logaM } ^{n} | = | a^{loga(Mn)} | Now, since a^{ (something) }= a^{(something else)}, then (something) = (something else), or in other words the exponents must be equal. | |
log_{a}M n | = | log_{a}(M^{n}) | And let's reverse the order of multiplication on left side... | |
n log_{a}M | = | log_{a}(M^{n}) | ...and that's it! |
Now, I assume textbooks will have it different still, and probably simpler... so I assume I didn't get quite there. Don't get me wrong, I have proved it! But I would need to try again, I guess, to find the ultimate simple looking proof.
But I hope you got the point how proving or any other mathematical problem solving is a process... and sometimes you may find a solution that is solution but not the most beautiful one. But see, if you are at the quest of trying to recreate those cookies you tasted at friend's house, even if you don't quite achieve the exact same taste or shape, you can still enjoy the cookies you made - if they still turn out to be cookies, that is.
I also realize that I did solve this problem fairly quickly and without astaying a lot, compared to what you would expect from a high school or college (I couldn't help it - after all I am a math teacher). But just remember, it's not bad to go astray for a while... as long as you keep "monitoring" your own progress and can turn back if it doesn't look too good - but at the same time don't "erase" all your erroneous work. Instead, analyze it. Analyzing your own work will so often give you a clue for a new route.
See also:
Dr. Math on logarithms - many many articles on logarithms and answers to students questions.