Precal Blog

Here are the slides on the problems we did today.

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I saw the most trouble with change of base. Let me give you the way to think about it one more time:

Say we are converting from base-2 to base-10. Look (click) at the diagram below and it should be clear that the unit distance (the distance between 0 and 1) on the base-10 scale is about 3 times as much as the unit distance on the base-2 scale.

So, if we knew exactly how many base-2 units it took to make a base-10 unit then we would have a conversion factor. Look closely again at the diagram. You should see that the base-10 scale is exactly `(log_10(10))/(log_10(2))` times as big. This, then is our conversion factor.

If the bases were equal our conversion would be `log_2(x)=log_10(x)`.

They are not equal so we apply the conversion factor and get: `log_2(x)= (log_10(10))/(log_10(2))*log_10(x)`.

This simplifies to give us the typical change of base formula: `log_2(x)=(log_10(x))/(log_10(2))`

This argument can be applied to the change of any base to another.

Here is a second area of trouble: Everyone is comfortable with `10^x=10000`. I think everyone is also ok with `log_10(10000)=x`. The first equation asks “What is the exponent I raise 10 to in order to get 10000″? The second equation asks the same question right? So when you see `log_3(x+1)=4` the question is “What is the exponent I raise 3 to in order to get `x+1`”? Write the question as an exponential equation: `3^a=(x+1)`. You know the answer because the logarithmic equation has a number after the equal sign. The logarithm(base 3 of `x+1`)is the exponent(4). Therefore, `3^4=x+1`.

A few of you asked for a printout of the last slide we looked at today, which I will do for you, but I also thought putting up some summary slides of what we have done may help you study and make sense of all the logarithm stuff we have been discussing.

There really isn’t an easy way out – it will take some time and meditation to understand this difficult concept.

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