Results 1 to 2 of 2

Thread: Logarithm problem with terminology I dont quite get

  1. #1
    Junior Member
    Jul 2012
    Bergen, Norway

    Logarithm problem with terminology I dont quite get

    Ok, this is a word problem in the book I am working in, and I am kind of stuck, because I dont completely understand all the terms, so I was looking for some guidance. The word problem is as follows:

    The intensity L(x) of light x meters under sea level is given as:


    where L0 = L(0) is the light intensity at sea level. A diver has found that the intensity is reduced to half at 6 meters under sea level. The diver cannot work without the aid of a lamp when the intensity is under 1/10 of the value of sea level. Calculate how deep the diver can go without the need for a lamp to complete his work.

    Ok, so I know that L(x) is just a function. However, I have not yet encountered the terms where they turn it into a subscript. (L0) So I thought to myself that since that has to equal 100% intensity, I can just change it out with 100, and then the problem becomes:


    The value I am looking for is 1/10 the value of 100, so 10. So the equation would be:


    However, there are two variables here. a and x. I understand what the x is, as it is how many meters down, and it is the variable that I am trying to solve. But what is with the a? I have a feeling that my lack of understanding of the terms here is locking me, and that the L0 shouldnt actually be set to 100, although it seems logical to me.

    Can anyone point me in the right direction? Thank you.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Jan 2008

    Re: Logarithm problem with terminology I dont quite get

    The a is a constant that tells us something about the clarity of the water.

    __________________________________________________ _________________________

    For example:

    With a=1 you would have $\displaystyle L(x)=L_0$, impossibly clear water.

    With a=1/2 you have $\displaystyle L(x)=L_0\left(\frac{1}{2}\right)^x$ or $\displaystyle L(x)=\frac {L_0}{2^x}$, for each metre you descend the light intensity is halved.

    __________________________________________________ _________________________

    In your question you are told that when x=6m the light intensity is half the surface intensity.

    Sub it in your equation $\displaystyle \frac{L(6)}{L_0}=\frac{1}{2}=a^6$.

    Can you solve that to find the value of a?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I dont know how to simplify this logarithm
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Oct 21st 2009, 12:22 PM
  2. word problem dont know were to start
    Posted in the Algebra Forum
    Replies: 5
    Last Post: Oct 17th 2008, 10:39 AM
  3. Replies: 12
    Last Post: Apr 6th 2008, 07:50 AM
  4. Replies: 2
    Last Post: Feb 20th 2008, 08:34 PM
  5. A terminology problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Jun 27th 2007, 08:11 AM

Search Tags

/mathhelpforum @mathhelpforum