Algebra 2 help =[
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thread starter DeeplyCool Feb 11 2010
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| I missed the class where we started to learn about logarithm. I went back the next day to get the homework. But I know nothing about logarithm.. I couldn't ask the teacher cuz I have the same classes every other day. It's due tomorrow for me.. =/ help please basilers? Here's one that I dont understand.. : log 12 = 1.1 log 8 = 0.9 log 7 = 0.8 Find log 7/8 So is it 0.8 divided by 0.9? What about: log 12 = 1.1 log 8 = 0.9 log 7 = 0.8 Find log 2/3 ... where does the 2/3 come from? |
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chaozd3mon
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| First one is pretty straight forward: Log 7 / Log 8 and the second one, you find the same ratio of 2/3 which would be 8/12 so therefore, log 8 / log 12 |
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Abba26
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 | Um, what is the question exactly? Those problems are already solved. log 12 = 1.1.... correct Are you supposed to show how to get those answers? Take 10^(1.1) and it equals approximately 12. That is how a common logarithm works. So if you have a problem like log 100 = x, you can rearrange it to look like this: 10^x = 100 So what does x have to be? 2. So log 100 = 2 The number you use for the base is always 10 if it just says log. Sometimes it will want you to use another number for the base in which case there will be a subscript after the "log" with a number to use for the base. EDIT: Oh, hold on a second. I see what they want you to do. For the first one there is a rule that says log (x/y) = log(x) - log(y) So log 7/8 = log(7) - log(8) = .8 -.9 = -.1 What the guy above me said is not quite right. For the next one, log 2/3 is the same thing as log 8/12 so you can do it the same way. log 8/12 = log(8) - log(12)... etc. |
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DeeplyCool 02/11/10
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My Characters: Sarahha - 9 Bellocan Magician
executionerz 02/11/10
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| What Chapter is this cuz were in chp 8.2 atm |
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