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Math Help - Logarithmic Functions....help =)

  1. #1
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    Logarithmic Functions....help =)

    Hey guys, having a little trouple with functions. Heres the problems:

    1) 4^x+2-4^x=15

    2)7^x^2=10

    3) 2^x=3^x

    and i have no idea how to post other q's with my keyboard

    thanx for the help though everyone
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  2. #2
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    Quote Originally Posted by Baginoman View Post
    Hey guys, having a little trouple with functions. Heres the problems:

    1) 4^x+2-4^x=15

    2)7^x^2=10

    3) 2^x=3^x

    and i have no idea how to post other q's with my keyboard

    thanx for the help though everyone
    1. I assume that you wanted to solve these equations(?)

    2. I assume that you meant:

    4^{x+2}-4^x=15~\implies~16 \cdot 4^x-4^x=15~\implies~ 15\cdot 4^x=15

    I leave the rest for you.

    7^{x^2} = 10~\implies~x^2=\log_7(10)~\implies~x=\sqrt{\log_7  (10)}

    2^x=3^x~\implies~1=\dfrac{3^x}{2^x} = \left(\dfrac32\right)^x

    Now use the definiton a^0=1\ ,\ a > 0

    to get x= 0
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  3. #3
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    awesome, those confused me lol

    how about:

    1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

    2) log(7y + 1)= 2 log(y + 3) - log 2

    3) log(3x - 4)/log x =2

    thank you again guys
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  4. #4
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    second question: just use logarithm rules,
    the 2 goes back to the exponent,then you have a difference of logarithms
    third question: change of bases rule
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  5. #5
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    Quote Originally Posted by Baginoman View Post
    ...

    1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

    2) log(7y + 1)= 2 log(y + 3) - log 2

    3) log(3x - 4)/log x =2

    ...
    1. Do us and do yourself a favour and start a new thread if you have new questions. Otherwise you risk that nobody will notice your need for help.

    2. I don't understand the equation in 1). Do you mean:

    (\log(3))^x or (\log_3)^x or \log(3^x) ...

    3. If you have some difficulties to use becker89 hint, here is a starter:

    \log(7y+1)=2\log(y+3)-\log(2)~\implies~ \log(7y+1)+\log(2)=2\log(y+3)~\implies~\log(14y+2)  =\log((y+3)^2)

    Now de-logarithmize. You'll get a quadratic equation in y. My result is: y = 7 or y = 1

    4. Multiply the equation through by log(x):

    \log(3x-4)=2\log(x)~\implies~\log(3x-4)=\log(x^2)

    De-logarithmize and solve the quadratic equation. This equation doesn't have a real solution.
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    Quote Originally Posted by earboth View Post
    1. Do us and do yourself a favour and start a new thread if you have new questions. Otherwise you risk that nobody will notice your need for help.

    2. I don't understand the equation in 1). Do you mean:

    (\log(3))^x or (\log_3)^x or \log(3^x) ...

    3. If you have some difficulties to use becker89 hint, here is a starter:

    \log(7y+1)=2\log(y+3)-\log(2)~\implies~ \log(7y+1)+\log(2)=2\log(y+3)~\implies~\log(14y+2)  =\log((y+3)^2)

    Now de-logarithmize. You'll get a quadratic equation in y. My result is: y = 7 or y = 1

    4. Multiply the equation through by log(x):

    \log(3x-4)=2\log(x)~\implies~\log(3x-4)=\log(x^2)

    De-logarithmize and solve the quadratic equation. This equation doesn't have a real solution.
    sorry about that, and yes the first one is (\log_3)^x

    also if its not a hassle, how does logX^2=2 come out as (10, -10)??

    P.s. do you mind if i PM you with some q's?
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  7. #7
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    Quote Originally Posted by Baginoman View Post
    sorry about that, and yes the first one is (\log_3)^x

    also if its not a hassle, how does logX^2=2 come out as (10, -10)??

    P.s. do you mind if i PM you with some q's?
    1. to (\log_3)^x: Now we know the base of the logarithmic function but the argument is still missing ...

    2. If \log means \log_{10} the the last equation can be solved:

    \log(x^2)=2~\implies~10^{\log(x^2)} = 10^2~\implies~x^2=100

    P.s. do you mind if i PM you with some q's?
    Don't do that: Give other members of MHF a chance !
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  8. #8
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    Quote Originally Posted by earboth View Post
    1. to (\log_3)^x: Now we know the base of the logarithmic function but the argument is still missing ...

    2. If \log means \log_{10} the the last equation can be solved:

    \log(x^2)=2~\implies~10^{\log(x^2)} = 10^2~\implies~x^2=100



    Don't do that: Give other members of MHF a chance !
    hahah thankyou very much, your talent makes me want to learn math to a greater extent!

    Plus i would post another thread for more help, but i kinda need an answer within the next half hour or so
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  9. #9
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    Quote Originally Posted by Baginoman View Post
    ...
    1) log3^x=log3(1/x)+4 (keep in mind both 3's are log of 3)

    ...
    I believe that I found the solution to this cryptic equation:

    Probably you mean:

    \log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~ 2\log_3(x)=4~\implies~x^2=81

    Since x > 0 there is only one solution. (x = 9)
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    Quote Originally Posted by earboth View Post
    I believe that I found the solution to this cryptic equation:

    Probably you mean:

    \log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~ 2\log_3(x)=4~\implies~x^2=81

    Since x > 0 there is only one solution. (x = 9)
    yep thats its lol. hmmm where did the 2log come from?
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  11. #11
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    Quote Originally Posted by Baginoman View Post
    yep thats its lol. hmmm where did the 2log come from?
    <br />
\log_3(x)=\log_3\left(\frac1x \right) + 4~\implies~\log_3(x)=-\log_3\left(x \right) + 4 ~\implies~<br />

    To get rid of the log at the RHS you have to add \log_3(x) on both sides of the equation. And then you'll get at the LHS:

    \log_3(x) + \log_3(x) = 2\log_3(x)
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