Posted: 3/25/2008 7:16:35 PM EDT
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Hi guys. I am reviewing in my Calc class and I found these two problems. I did them, but they don't look right to me at all.... What do you guys get: Solve for X 50/(4+e^2x)=1 ln(3x+4)-ln(2x+1)=5 ETA: I am on the phone with a friend in my class and he has way different answers - But he's not sure if they are right either. The problem is that I cannot find a good example anywhere... 
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The first one, cross multiply both sides to get 50 = 4 +e^2x subtract 4 to get your e term isolated take the natural log of both sides divide by two The second, use properties of logs to combine the difference into a single quotient, then exponentiate both sides to drop the ln and solve using algebra All hail the Napier! |
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To get my degree I slogged through 5 semesters of calculus, a few prob/stats classes and an introduction to combinatorics. 6 years later, I barely remember the basics of integration. After I completed the "pure math" classes, I used a calculator for almost everything. I found the machine to be faster and more accurate than I was. If you can quickly pick up RPN, an HP calc is the way to go. Otherwise, a TI-89 will serve you well. |
I have been in the engineering field for two years... I used it ONCE. My x-boss asked me as a joke to come up with a formula for the radius on the arch over his doorway, and solve it - hell it was overtime so I did it 
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That still isn't calculus. 
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This is algebra It's not calculus unless there are derivatives or integrals. 50/(4+e^2x)=1 50 = 4+e^2x 46 = e^2x ln46 = ln e^2x ln46 = 2x (ln46)/2 = x the next one... Check this... I'm not sure if I'm right. Look up properties of ln. ln(3x+4)-ln(2x+1)=5 ln ( (3x+4) / (2x + 1) ) = 5 e^ln ( (3x+4) / (2x + 1) ) = e^5 (3x+4) / (2x + 1) = e^5 3x +4 = e^5 (2x + 1) 3x +4 = 2e^5 x + e^5 0 = 2e^5 x -3x + e^5 -4 -e^5 +4 = 2e^5 x -3x (-e^5 +4)/(2e^5 - 3) = x Again check this... EDIT: 50/(4+e^2x)=11 50 = 11(4+e^2x) 6 = 11e^2x 6/11 = e^2x ln6/11 = ln e^2x ln6/11 = 2x (ln6/11)/2 = x |
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Both of Master_of_Orion's answers are correct. You might be in trouble in your calc class if you cant do those. for the first one just remember that taking the natural log of both sides will bring down the exponent of e. ex. 12= e^x take ln of both sides ln(12) = x for the second one you need to realize that subtracting two ln's is the same as taking the natural log of their fraction. ex. ln(5) - ln(3) = ln (5/3) actually the algebra part of Calculus is usually the hardest part, so good luck. |
