Base e, In, LogDate: 03/08/2002 at 05:28:55 From: Jerry Harper Subject: Base e, In, Log First of all, I would like to thank you for telling us, how the value of base e works. I have some problems below, in base e. I've tried everything, but can't come up with the same answer as the book, and they don't give a clear example, or next to none of how they got their answer. Would you give me the complete clue on how they got their answer please? 3e^(2x-1) = 7 the textbook answer is 1.079 I have tried: Div^Log(7+3) 2Log3 = 1.047.... Div^(log7+log3) 2Log3 = 1.3856 Div^(Log(e)7+Log(e)3) 2Log(e)3 = 1.666... e^(x+1) = 8 the textbook answer is .924 I tried a similar approach, but no luck! We very much appreciate your site! Date: 03/08/2002 at 09:07:17 From: Doctor Peterson Subject: Re: Base e, In, Log Hi, Jerry. Thanks for your comments! We're always happy to see people both reading our site and asking questions. You solve this kind of problem just the way you solve other basic algebra problems: undo what is done on the left, one step at a time, until you get the variable alone. Note what is being done: 3 e^(2x-1) = 7 If you knew the value of x, you would 1. multiply by 2 2. subtract 1 3. raise e to that power 4. multiply by 3 and you'd get 7. How can we undo this? Think about putting on and taking off your shoes: in the morning, you first put on your socks, then your shoes; in the evening you take off your shoes, then your socks. You undo each step in the reverse order. Same here; undo each step this way: 4. divide by 3 3. take the natural log 2. add 1 1. divide by 2 Here we go: 3 e^(2x-1) = 7 e^(2x-1) = 7/3 2x-1 = ln(7/3) 2x = ln(7/3) + 1 x = [ln(7/3) + 1]/2 = [ln(2.3...] + 1]/2 = [0.8473 + 1]/2 = 1.8473/2 = 0.9236 I think you must have copied the wrong answers! Of course, you can (and always should) check your answer by plugging it back in: 3 e^(2*0.9236-1) = 3e^0.8473 = 3*2.333... = 7 See if you can do the other problem the same way. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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