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Homework Help: Algebra 2 natural logarithms problem

  1. Apr 16, 2004 #1
    Problem 51.
    At a constant temperature, the atmospheric pressure p, in pascals, is given by the formula p=101.3e^-0.001h, where h is the altitude in meters. Find h when p is 74.3 pascals. Hint: Start by taking the natural logarithms of the expressions on each side of the equation.

    Solve and check.
    41. (29.3)^4x-1=(17.3)^5x+2
     
  2. jcsd
  3. Apr 16, 2004 #2
    1. The problem tells you how to solve it. What've you tried?

    2. Manipulate the expression to get a common exponent, and then take the logarithm. What've you tried so far?

    cookiemonster
     
  4. Apr 16, 2004 #3
    regard on problem 51.

    So it would be:
    p=74.3
    p=101.3e^-0.001h
    74.3=101.3e^-0.001h
    74.3=274.3619^-0.001
    Is this right, so far? If so what is the next step??
     
  5. Apr 16, 2004 #4

    enigma

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    What did you do to go through that step?

    a latural logarithm is like "the opposite" of an e^

    If you put them together, you get 1.

    So, take ln of both sides to get rid of the e^-0.001*h, and give you *some number* times -.001h on the right side, and ln(74.3) on the left.
     
  6. Apr 16, 2004 #5
    What'd you do between the third and fourth line? How'd you make the h disappear? Why wasn't the first step you did taking the natural logarithm of both sides like the problem says?

    cookiemonster
     
  7. Apr 16, 2004 #6
    Cont...

    p=74.3
    p=101.3e^-0.001h
    74.3=101.3e^-0.001h
    ln(74.30)=ln(101.3)*-0.001h
    4.3081=4.618086*-0.001h
    Is this right so far??

    Any given program, when running, is obsolete.
    (Laws of Computer Programming, I)
     
  8. Apr 16, 2004 #7
    Careful.

    ln(101.3*e^(-.001h)) = ln(101.3) + ln(e^(-.001h)) = ln(101.3) - .001h

    cookiemonster
     
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