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Math Help - logarithms and word problems

  1. #1
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    logarithms and word problems

    THANK YOU GUYS SO MUCH FOR HELPING WITH THE PROBLEMS THAT I RECENTLY POSTED, SO IF IT'S NOT A PROBLEM I'M POSTING A FEW MORE THAT I COULD USE SOME ADVICE ON..

    log(x^3) + log (y^2) + log (y) + log(z) + log(z) + log (z)


    log_3(x^2 - 3x - 1)= log_3(5)


    2log(x) = log(12-2x)


    ln(x) - ln(3)= 2ln(x+1)


    3ln(x+3) - ln(x-3) =2ln(x-1)


    ^those are the new problems, please help! thankss!
    -----------------------------------------------------------------------------------------------


    Received a take-home test for my college algebra class.. stuck on a word problem..

    A radioactive material has a half-life of 3 days. Its level of radioactivity is measured at 1,000,000 units. 10 units is dangerous. How long will it take for the level of radioactivity to decrease to just one unit? To .001 unit?



    also..

    ln(2x/5- 3)= ln(2 + x/3)

    one more..

    A pnd begins to develp a cover of lily pads. The number of lily pads doubles every 3 days. after 49 days, the pond is completely covered. How long before the pond is one quarter covered? Assume the lily pads are all the same size.





    any help would be wonderful!
    Last edited by brittany12; April 30th 2007 at 10:55 AM.
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  2. #2
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    Hello, Brittany!

    The lily-pad problem is a trick question.
    . . And we have a trick answer for it . . .


    A pond begins to develop a cover of lily pads.
    The number of lily pads doubles every 3 days.
    After 49 days, the pond is completely covered.
    How long before the pond is one-quarter covered?
    Assume the lily pads are all the same size.

    "The number of lily pads doubles every 3 days."

    The pond is completely covered on the 49th day.

    This means that 3 days ago (46th day), the pond was half covered.

    And 3 days before that, the pond was one-quarter covered.

    Answer: .43 days.

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  3. #3
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    that makes everything much clearer now! i just need to write it out into an equation and that solves that problem! thanks!
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by brittany12 View Post
    Received a take-home test for my college algebra class.. stuck on a word problem..

    A radioactive material has a half-life of 3 days. Its level of radioactivity is measured at 1,000,000 units. 10 units is dangerous. How long will it take for the level of radioactivity to decrease to just one unit? To .001 unit?
    now there is a formula that says, half-life = ln2/r but we could derive this if you prefer.

    Now, since half-life = 3 = ln2/r
    => 3 = ln2/r
    => r = ln2/3 = 0.23104906

    Now we use the formula,

    A(t) = Ao*e^(-rt)
    where A(t) is the amount at time t, Ao is the initial amount, r is the rate of decrease, and t is the current time.

    So, we have Ao = 1000000, we want A(t) = 1

    So, 1 = 1000000*e^(-0.23104906t)
    => 0.000001 = e^-0.23104906t
    => ln0.000001 = -0.23104906t
    => t = ln0.000001/-0.23104906 = 59.79 days

    You can try the one for 0.1 units, it's the same procedure.






    ln(2x/5- 3)= ln(2 + x/3)
    I suppose you want to solve for x

    ln(2x/5 - 3) = ln(2 + x/3)
    since we have the log of something equals the log of something else, then something = something else

    => 2x/5 - 3 = 2 + x/3
    => 2x/5 - x/3 = 5
    => (6x - 5x)/15 = 5
    => x = 75
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  5. #5
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    amazing, thank you sooo so so so so so much!
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Two comments before we begin:
    1) When you have new questions, please post them in a new thread, it is for your own good. When you edit a post, it does not show up on the system as a new post, it just gets left in the stack wherever it is. further more, those who are involved in the post would not recognize a difference, since the same number of posts are in the thread as the last time they left. at the very least, you should put new questions in a new post that way, the thread would be identified as a new post and placed in an area for everyone to see.

    2) Please state clearly what your questions are. you type math very well, understanding what you typed is not the problem, the problem is you neglect to give the instructions for your questions, and so whoever's answering them would have to guess what they should do. here it's not so bad, but it could be

    Quote Originally Posted by brittany12 View Post
    log(x^3) + log (y^2) + log (y) + log(z) + log(z) + log (z)
    i assume you want to write this as a single log, again, just guessing

    log(x^3) + log (y^2) + log (y) + log(z) + log(z) + log (z)

    = log(x^3) + log (y^2) + log (y) + 3log(z) ........added all the log(z)'s
    = log(x^3) + log (y^2) + log (y) + log(z^3) ......changed 3log(z) to log(z^3) using a law of logarithms
    = log[(x^3)(y^2)(y)(z^3)] .............................since in general, log(a) + log(b) = log(ab)
    = log[(x^3)(y^3)(z^3)]


    log_3(x^2 - 3x - 1)= log_3(5)
    I suppose you want to solve for x, we have log_3 on both sides, we can just drop the logs then

    log_3(x^2 - 3x - 1) = log_3(5)
    => x^2 - 3x - 1 = 5
    => x^2 - 3x - 6 = 0 .........this is weird, are you sure it wasn't a + 1 on the left?
    anyway, by the quadratic formula,
    x = [3 +/- sqrt(9 + 24)]/2
    => x = [3 +/- sqrt(33)]/2
    => x = [3 + sqrt(33)]/2 or [3 - sqrt(33)]/2


    2log(x) = log(12-2x)
    Solve for x
    2log(x) = log(12-2x)
    => log(x^2) = log(12 - 2x) ..............since log(x^n) = nlog(x)
    => x^2 = 12 - 2x
    => x^2 + 2x - 12 = 0
    By the quadratic formula
    x = [-2 +/- sqrt(4 + 48)]/2
    => x = [-2 +/- sqrt(52)]/2
    => x = [-2 + sqrt(52)]/2 only, since we can't have negative x

    ln(x) - ln(3)= 2ln(x+1)
    Solve for x

    ln(x) - ln(3) = 2ln(x+1)
    => ln(x/3) = ln(x + 1)^2 ...........since logx - logy = log(x/y) and nlogx = log(x^n)
    => x/3 = (x + 1)^2
    => x/3 = x^2 + 2x + 1
    => x = 3x^2 + 6x + 3
    => 3x^2 + 5x + 3 = 0
    By the quadratic formula:
    x = [-5 +/- sqrt(25 - 36)]/2
    no solution, the two sides are never equal

    3ln(x+3) - ln(x-3) = 2ln(x-1)
    Solve for x
    3ln(x+3) - ln(x-3) = 2ln(x-1)
    => ln(x + 3)^3 - ln(x - 3) = ln(x - 1)^2 ...........since nlogx = log(x^n)
    => ln[{(x + 3)^3}/(x - 3)] = ln(x - 1)^2 ..........since logx - logy = log(x/y)
    => [(x + 3)^3]/(x - 3) = (x - 1)^2
    => (x + 3)^3 = (x - 3)(x - 1)^2
    => x^3 + 9x^2 + 27x + 27 = (x - 3)(x^2 - 2x + 1)
    => x^3 + 9x^2 + 27x + 27 = x^3 - 5x^2 + 7x - 3
    => 14x^2 + 20x + 30 = 0
    => 7x^2 + 10x + 15 = 0
    By the quadratic formula:
    x = [-10 +/- sqrt(100 - 420)]/14
    again no solution

    yup, you have some weird log problems
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