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Math Help - I Am Struggling A Little Bit - Logarithmic Equations

  1. #1
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    I Am Struggling A Little Bit - Logarithmic Equations

    I am really having trouble with logarithms.

    How would I go about solving these equations?

    I am struggling quite a bit with logarithms so to be honest I am not really sure where to begin. I know I must get all of the logs on one side and numbers on the other, but that is about it.

    1) y=200e^(0.05t)+50 solve for t

    For number 1 would I begin by moving the 50 to the other side so I would have y-50= 200e^(0.05)
    Would I now take the natural log to both sides?

    2) 10log(I/10-12) = 87 solve for I

    For this one I believe I would begin with taking 10 to both sides to cancel out the log.

    So would I have 10(I/10^-12) = 10^87? Is this correct so far? Where would I go from here?

    3) y = 5log base 2 of (x-3)-4 solve for X

    For this one I know I need to isolate x. So would I begin by adding 4 to both sides?
    If I did that I would have y+4 = 5log base 2 of (x-3). From here I am not quite sure which direction to head.


    So as you can see I am struggling quite a bit. But I am trying. I would appreciate any and all help I can get. Thank you!
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  2. #2
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    Quote Originally Posted by Adampff View Post
    I am really having trouble with logarithms.

    How would I go about solving these equations?

    I am struggling quite a bit with logarithms so to be honest I am not really sure where to begin. I know I must get all of the logs on one side and numbers on the other, but that is about it.

    1) y=200e^(0.05t)+50 solve for t

    For number 1 would I begin by moving the 50 to the other side so I would have y-50= 200e^(0.05)
    Would I now take the natural log to both sides?

    divide both sides by 200 first, then take the natural log of both sides

    2) 10log(I/10-12) = 87 solve for I

    For this one I believe I would begin with taking 10 to both sides to cancel out the log. no, that is not correct.

    10log(I/10-12) = 87

    divide both sides by 10 first ...

    log(I/10-12) = 8.7

    (I/10 - 12) = 10^8.7


    3) y = 5log base 2 of (x-3)-4 solve for X

    For this one I know I need to isolate x. So would I begin by adding 4 to both sides?
    If I did that I would have y+4 = 5log base 2 of (x-3). From here I am not quite sure which direction to head.

    (y+4)/5 = log_2(x-3)

    2^[(y+4)/5] = x-3

    ...
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  3. #3
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    Alright, thank you very much for the help!

    For number 1 my final answer is...

    ln(y-50/200/0.05) = t My teacher said it was ok right right it like this.

    For number 2 my final answer is...

    I = 10^-3.0

    I got this answer like this...
    10log(I/10^-12) = 87 original problem
    log(I/10^-12) = 8.7 divided by 10
    I/(10^-12) = 10^8.7 took 10 to both sides to get rid of log
    I = 10^-3.3 multiplied by 10^-12 to both sides and added exponent

    As I am sure you can see I did actually right the original problem incorrect for this one. I originally had written out my problem like this. 10log(I/10-12) = 87 solve for I, but I forgot the carrot symbol between the 10 and -12. It should actually look like this. 10log(I/10^-12) = 87 solve for I

    I am sorry about this.

    As for number 3 my final answer is...

    2^[y+(4/5)] + 3 = x

    I just added 3 to both answers from where you left off in your post.

    I am extremely happy for you help. Math is certainly something I struggle with and you explained things perfectly. Thank you very much!
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