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Math Help - logs and powers questions

  1. #1
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    logs and powers questions

    Hi I'm new here. Unfortunately I dont have the time to make an official introduction, since I've been scratching my head over some assignments for the past three days and now I only have two hours left to do them.

    I'd be really grateful if someone could help me with the following (listed from easiest to most difficult):

    1) 2,2= -log(c), 2,2= pH value and c= molar concentration.

    What is c in the above equation?

    If the molar concentration in a liquid is multiplied by 100, the pH will be reduced by 2. Explain why (something to do with logarithmic rules).


    2) f(x)= 225.027* x^ -0.5

    - What is x when f(x) is 10.000?
    - By which percent-rate will x increase if y decreases with 5 %

    Now they're not that difficult, but I think the reason I can't seem to solve them is because I'm not sure if the reverse function of a negative power in this case x^-0.5, is 1/sqr(-0.5). When I try to isolate x, it certainly doesn't work to use 1/sqr(-0.5) in the process. Nor does it help when I try to use it to calculate the percent rate increase in x, for which the formula is: 1+ry= (1+rx)^a, rx= percent rate for x, ry= percent rate for y.


    Please help ASAP!

    Thank you in advance.
    Last edited by CaptainBlack; March 21st 2011 at 09:26 AM.
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  2. #2
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    Quote Originally Posted by Lia85 View Post
    Hi I'm new here. Unfortunately I dont have the time to make an official introduction, since I've been scratching my head over some assignments for the past three days and now I only have two hours left to do them.

    I'd be really grateful if someone could help me with the following (listed from easiest to most difficult):

    1) 2,2= -log(c), 2,2= pH value and c= molar concentration.

    What is c in the above equation?
    The "inverse" of log (base 10) is 10 to a power. Write that as log(c)= -2.2 and take 10^{log(c)}= 10^{-2.2}.

    If the molar concentration in a liquid is multiplied by 100, the pH will be reduced by 2. Explain why (something to do with logarithmic rules).
    100= 10^2 so that log(100)= 2.


    2) f(x)= 225.027* x^ -0.5

    - What is x when f(x) is 10.000?
    That's the same as \frac{225.027}{\sqrt{x}}= 10.00 or \sqrt{x}= \frac{10}{225.027}

    - By which percent-rate will x increase if y decreases with 5 %
    Do you mean y= f(x)? 10.000 decreased by 5% will be 9.500. Solve the same equation with 9.5 instead of 10 to get the new value of x. The percentage increase will be "(new x- old x)/old x.

    [Now they're not that difficult, but I think the reason I can't seem to solve them is because I'm not sure if the reverse function of a negative power in this case x^-0.5, is 1/sqr(-0.5).
    NO, it is just 1/sqrt(0.5). The negative power puts the square root in the denominator but does not change a sign.

    When I try to isolate x, it certainly doesn't work to use 1/sqr(-0.5) in the process. Nor does it help when I try to use it to calculate the percent rate increase in x, for which the formula is: 1+ry= (1+rx)^a, rx= percent rate for x, ry= percent rate for y.


    Please help ASAP!

    Thank you in advance.
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  3. #3
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    Hi HallsosIvy and thank you for your quick reply.


    "Do you mean y= f(x)? 10.000 decreased by 5% will be 9.500. Solve the same equation with 9.5 instead of 10 to get the new value of x. The percentage increase will be "(new x- old x)/old x. "

    Yes I know this, but because this is a power function, we have to use this formula 1+ry= (1+rx)^a, rx= percent rate for x, ry= percent rate for y, in order to explain the relative increment in x (i.e. the percent rate by which it increases when y decreases). And when I try this I get awfully weird numbers (that I didn't even know existed), because I can't seem to properly isolate rx in that equation (it should be a bumber between one and zero).

    Again thank you
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  4. #4
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    nvm
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