2007 paul dawkins 302

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Unformatted text preview: of zero. Next, in order to move the exponent down it has to be on the whole term inside the logarithm and that just won’t be the case with this equation in its present form. So, the first step is to move on of the terms to the other side of the equal sign, then we will take the logarithm of both sides using the natural logarithm. 2 4 y +1 = 3 y ln 2 4 y +1 = ln 3 y ( 4 y + 1) ln 2 = y ln 3 Okay, this looks messy, but again, it’s really not that bad. Let’s look at the following equation first. 2 ( 4 y + 1) = 3 y 8y + 2 = 3y 5 y = -2 2 y=5 We can all solve this equation and so that means that we can solve the one that we’ve got. Again the ln2 and ln3 are just numbers and so the process is exactly the same. The answer will be messier than this equation, but the process is identical. Here is the work for this one. ( 4 y + 1) ln 2 = y ln 3 4 y ln 2 + ln 2 = y ln 3 4 y ln 2 - y ln 3 = - ln 2 y ( 4 ln 2 - ln 3) = - ln 2 y=- ln 2 4 ln 2 - ln 3 So, we get all the terms with y in them on one side and a...
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This note was uploaded on 06/06/2012 for the course ICT 4 taught by Professor Mrvinh during the Spring '12 term at Hanoi University of Technology.

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