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Math Help - Natural Logarithums

  1. #1
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    Question Natural Logarithums

    HI....Sorry to bother you all. I posted on here before with a problem I have been having. As I am new here I am getting to grips with using the great site and getting some help from people in the know. Please find the thumb nail attached. I thought I knew what I was doing but have really lost me way. If some one could lay this out so I could understand that would be very helpful.
    Attached Thumbnails Attached Thumbnails Natural Logarithums-help-natural-logs.jpg  
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  2. #2
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    Re: Natural Logarithums

    Quote Originally Posted by camorris View Post
    HI....Sorry to bother you all. I posted on here before with a problem I have been having. As I am new here I am getting to grips with using the great site and getting some help from people in the know. Please find the thumb nail attached. I thought I knew what I was doing but have really lost me way. If some one could lay this out so I could understand that would be very helpful.
    L = {L_0}\left( {{e^{h\omega }} + 1} \right)
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  3. #3
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    Re: Natural Logarithums

    So the problem is: given \omega= \frac{1}{h} ln\left(\frac{L}{L_0}-1\right), find L when \omega= -2.6, L_0= 16 and h= 1.5.

    Putting in the given values, -2.6= \frac{1}{1.5}ln\left(\frac{L}{16}- 1\right).

    Notice what that says- if you divide L by 16, then subtract 1, take the natural logarithm, and divide by 1.5, then the result is -2.6. To go the opposite way, from -2.6 to a value of L, you have to do the opposite- do the opposite of each operation, in the opposite order. Since the last thing done above was "divide by 1.5", the first thing we should do to solve for L is "multiply by 1.5". Doing that on both sides gives -2.6(1.5)= -3.9= ln\left(\frac{L}{16}- 1\right). The opposite of "natural logarithm", ln, is its "inverse function" the exponential. Taking the exponential of both sides gives e^{-3.9}= 0.0202= \frac{L}{16}- 1. The opposite of "subtract 1", -1, is "add 1". Adding 1 to both sides, 0.0202+ 1= 1.0202= \frac{L}{16}. Finally, the opposite of "divide by 16" is, of course, "multiply by 16". Multiplying both sides by 16, (0.0202)(16)= 0.3239= L
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  4. #4
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    Re: Natural Logarithums

    Thank you so much. I did it the first time and got the same format as Plato has provided although I did the (e^hw - 1).

    I see I have to formats to go from and to different ways. Thank you very much once again.
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