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Properties of Logarithms - Problem 1
PhD. in Mathematics
Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.
I want to do some examples that are going to illustrate the use of the properties of logarithms. Let's review those really quickly. They are three. We've got the log of a product, log base b of x times y, equals log base b of x, plus log base b of y. The log of a quotient; log base b of x, over y equals log base b of x, minus log base b of y. The log of a power; log base b of x to the n equals n times log base b of x.
This is like pulling the exponents outside. So here is the problem. Suppose you know that the log of 2, and this is the common log, I just haven't written a base here, so this is assumed to be log base 10. The common log of 2 is about.3, which it is. Use this to evaluate the following.
So one of the really neat about logarithms is that, if you know one or two values, you can actually use them to figure out other values. Well the first example doesn't actually use this fact. The log of 10 to the n. Since this is just the common log, it's log base 10, we have the log base 10 of 10 to the n. By an inverse property of logarithms, this is just n. I have this here, because it's going to be a useful fact in the rest of the examples.
Log of 200. So in order to use one of the properties, I have to express this somehow in terms of 2. Well, 200 is 2 times 100, and by the log of a product property, this is log of 2, plus the log of 100. Now the log of 2 is approximately .3, and the log of 100 by this property is 2. So this is 2.3.
The log of 50. Again I want to try to express 50 in terms of 2 and powers of 10. 50 is 100 divided by 2. So I've got the log of 100 minus the log of 2 by the log of a quotient property. Log of a quotient is the difference of logs. Log 2 is approximately.3, so this is approximately 2 minus .3, and that's 1.7.
Log of 500, we can use the fact we just found that log of 50 is 1.7 here and write this as log of 500 times 10. So this is log of 50, plus log of 10. I should make an approximation here. This is approximately equal to, because this 1.7 is an approximation for log of 50. 1.7 plus log of 10 is 1, 2.7.
Log of .125. So I have to find some way to express this in terms of 2's and 10's. .125 is 1/8. 1/8 is 2 to the -3. By the power of property, log of a power, this is -3, times the log of 2. The log of 2 is approximately .3. So this is minus 3 times .3, minus .9.
Finally, log of 3.2. The log of 3.2, this one is tough, but I know that 32 is a power of 2. So this is the same as log of 32 over 10. So this becomes log of 32, minus log of 10. The log of 32, that's the log of 2 to the 5th. Log of 10 is 1. Log of 2 to the 5th equals 5, times log of 2, I carried over the minus 1, and again the log of 2 is approximately .3. 5 times .3 minus 1. 5 times .3, 1.5. So the log of 3.2 is approximately .5.
Do you know what that means? If you raise 10 to the .5, you would get 3.21. So 3.2 is approximately the square root of 10.
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