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Math Help - more problems I dont get.

  1. #1
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    more problems I dont get.

    more problems I can't understand

    9^x+2 = 27^-3

    I get that 3 will go into both of them but am lost after that, notes don't help

    instructions are to solve each exponential equation. Round to 2 decimal places.


    and another logarithm

    Log (xy^2/Z^4)
    Z


    edit: Z should be right after/underneath "log" and before the (.
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  2. #2
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    Quote Originally Posted by kenny87 View Post
    more problems I can't understand

    9^x+2 = 27^-3
    9^{x+2} = 27^{-3}

    (3^2)^{x+2} = (3^3)^{-3}

    3^{2(x+2)} = 3^{3\times -3}

    2(x+2) = 3\times -3

    Can you finish this?
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  3. #3
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    I guess the problem says 9^(X+2)=27^-3. In this case, (3^2)^(X+2)=(3^3)^-3 and get 3^(2X+4)=3^-9 and from that point you can go on.
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  4. #4
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    Quote Originally Posted by kenny87 View Post
    and another logarithm

    Log (xy^2/Z^4)
    Z

    edit: Z should be right after/underneath "log" and before the (.
    What are you supposed to do with this expression?

     <br />
\log_z \frac{xy^2}{z^4} = \log_z xy^2- \log_z z^4 = \log_z xy^2- 4\log_z z = \log_z xy^2- 4\times 1= \log_z xy^2- 4<br />
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  5. #5
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    says to expand or condense expression as much as possible.
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  6. #6
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    more problems I can't understand

    9^x+2 = 27^-3

    my bad, problem is 9^x+2 = 27^ -x


    I must have been looking two places at once or something. I figured that one out though looking at what you did.

    It was in the first section of my notes, one more though

    e^5x =77

    I have one example dealing with E but it has E on both sides so not sure what to do here. Hopefully this is my last question, thanks for the help.
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  7. #7
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    Quote Originally Posted by kenny87 View Post

    , one more though

    e^5x =77
     e^{5x} =77

     5x =\ln 77

     x =\frac{\ln 77}{5}

    where \ln = \log_e
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  8. #8
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    thanks, so I guess I would take e * 77 and divide by 5 to get my decimal answer?
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  9. #9
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    Quote Originally Posted by kenny87 View Post
    thanks, so I guess I would take e * 77 and divide by 5 to get my decimal answer?

    No, you would find \frac{\log_e77}{5} to get a decimal approximation.
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