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Math Help - 1 Word problem and 1 function problem

  1. #1
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    1 Word problem and 1 function problem

    you guys are all very good at math, i love these forums. I understand more things here then i do most the time in class, i have a test coming up and i really need help understanding these 2 problems, can someone please solve these problems and explain the steps, i would really appreciate it.


    i got a review packet, i scanned the problems because i dont know how to type them
    http://img694.imageshack.us/img694/9...wrk2222222.jpg

    problem #28 and #29 are the ones im having problem with and need the help.

    also was i correct with #27? was it A? thanks everyone
    Last edited by mr fantastic; April 20th 2010 at 08:23 PM. Reason: Edited post title
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  2. #2
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    28) For 2^x = 15

    consider the rule when a^b=c \implies b = \log_ac this will give you the answer

    29) And for A= 456 in A= 282 e^{0.04t} then

    456= 282 e^{0.04t} solve for t
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  3. #3
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    27) Your answer is incorrect. If you use a calculator with your answer as x it doesn't solve the equation.

    on both sides of the equation use the following
    a^x=b^y \Longrightarrow \ln (a^x)=\ln(b^y) \Longrightarrow x\ln(a)=y\ln(b)

    you may need to use

    \log_ac=\frac{\log_{10}c}{\log_{10}a} if you need to change base to use your calculator.

    Edit: My bad you don't need to evaluate \log_ac but worth knowing anyway :s
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  4. #4
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    Quote Originally Posted by Krahl View Post
    27) Your answer is incorrect. If you use a calculator with your answer as x it doesn't solve the equation.

    on both sides of the equation use the following
    a^x=b^y \Longrightarrow \ln (a^x)=\ln(b^y) \Longrightarrow x\ln(a)=y\ln(b)

    you may need to use

    \log_ac=\frac{\log_{10}c}{\log_{10}a} if you need to change base to use your calculator.

    Edit: My bad you don't need to evaluate \log_ac but worth knowing anyway :s
    i have spent this last hour trying to figure out #27. can you please solve this problem, is it B or C?
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  5. #5
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    Quote Originally Posted by murkr View Post
    i have spent this last hour trying to figure out #27. can you please solve this problem, is it B or C?
    For #27
    xln(6/7) = (1-x)ln(2)
    Now solve for x.
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  6. #6
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    Quote Originally Posted by sa-ri-ga-ma View Post
    For #27
    xln(6/7) = (1-x)ln(2)
    Now solve for x.
    yes i got to the next step

    x * (ln(6/7) + ln[2]) = ln[2]

    can you please work the question out in steps, i understand it a lot more when i see it done in steps, thanks for all your help. and actually trying to get me to figure it out myself and not just give me the answer haha. but im really having a hard time and my test is tomrrow so please solve this in steps for me, thank you.
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  7. #7
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    x * (ln(6/7) + ln[2]) = ln[2]

    One last step for the result.
    If ax= b, then
    x = .....?
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  8. #8
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    Quote Originally Posted by sa-ri-ga-ma View Post
    x * (ln(6/7) + ln[2]) = ln[2]

    One last step for the result.
    If ax= b, then
    x = .....?
    is it C?? thats what i got, if its wrong im stumped please finish.. i just forgot how to do this
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  9. #9
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    \frac{6}{7}^x= (2)^{1-x}

    take log of both sides

    ln\frac{6}{7}^x= ln(2)^{1-x}

    the exponent comes down

    xln\frac{6}{7}= (1-x)ln(2)

    Collect like terms

    xln\frac{6}{7}= 1ln(2)-xln(2)

    xln\frac{6}{7}+xln(2)= 1ln(2)

    x(ln\frac{6}{7}+ln(2))= 1ln(2)

    and divide both sides by ln\frac{6}{7}+ln(2)) and yes you get C
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