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Math Help - Some math problems i cant seem to figure out? Part 3

  1. #1
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    Unhappy Some math problems i cant seem to figure out? Part 3

    1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years

    2. Write in logarithmic form: 2^-2=1/4?


    3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?



    4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4
    Last edited by mr fantastic; January 5th 2009 at 09:38 PM.
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  2. #2
    Junior Member shinhidora's Avatar
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    Quote Originally Posted by rotw121 View Post
    1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years

    2. Write in logarithmic form: 2^-2=1/4?


    3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?



    4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4
    I myself can't really help you with logarithms I utterly, you know what, at them but I've taken the liberty to put em in LaTex for ya as I know some of the members don't like solving stuff that isn't put in a nice visual way hope this speeds up getting an answer and so I atleast help you in a certain way

    2. 2^{-2}=\frac{1}{4}

    3. \log_{3} \frac{1}{9} = -2

    4.
    a. \log_{3} x=2

    b. \log_{3} (x-1)=4
    Last edited by mr fantastic; January 6th 2009 at 09:17 AM. Reason: Fixed the latex.
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  3. #3
    Super Member craig's Avatar
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    2. 2^{-2}=\frac{1}{4}
    \log_{2} {\frac{1}{4}}= -2

    Sorry but my minds gone blank about the rest.
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  4. #4
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    Quote Originally Posted by rotw121 View Post
    1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years

    2. Write in logarithmic form: 2^-2=1/4?


    3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?



    4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4
    Quote Originally Posted by shinhidora View Post
    [snip]
    2. 2^{-2}=\frac{1}{4}

    3. \log_{3} \frac{1}{9} = -2

    4.
    a. \log_{3} x=2

    b. \log_{3} (x-1)=4
    Q3. and Q4. are done by applying the fact that \log_A B = C \Rightarrow A^C = B.

    For Q1, perhaps you could first tell us what the basic compound interest formula is .....
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  5. #5
    A riddle wrapped in an enigma
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    Quote Originally Posted by rotw121 View Post
    1. use the compound interest formula to determine the final value of the given amount: $1,000 at 8% compounded annually for 3 years

    2. Write in logarithmic form: 2^-2=1/4?


    3. Write an equivalent expression in exponential form: log lower case3 to (1/9)=-2?



    4. Solve the equation: 1. log lower case 3 to x=2 2. log lower case 3 to (x-1)=4

    Hello rotw,

    [1.] Compound interest formula:

    A=P\left(1+\frac{r}{n}\right)^{nt}

    A=1000\left(1+\frac{.08}{1}\right)^{3}

    Easy now....

    [2.] 2^{-2}=\frac{1}{4} \Longrightarrow \log_2 \frac{1}{4}=-2

    [3.] \log_3 \frac{1}{9}=-2 \Longrightarrow 3^{-2}=\frac{1}{9}

    [4.1] \log_3 x=2

    x=3^2


    [4.2] \log_3(x-1)=4

    3^4=x-1
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  6. #6
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    hey sorry guys i figured it out already but Thanks everyone for helping me out
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