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Math Help - logs and exponentials question

  1. #1
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    Question logs and exponentials question

    solve, giving your answers as exact fractions, the simultaneous equations:

    8^(x) = 4^(2x+3)

    log(base2)y = log(base2)x + 4

    ('^' meaning 'to the power of')
    I am stuck, i dont know how to do it! PLease show your method,
    Thanks, azza xxxz
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  2. #2
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    Quote Originally Posted by azza View Post
    solve, giving your answers as exact fractions, the simultaneous equations: 8^(x) = 4^{2x+3}
    8^x=4^{2x+3} is the same as 2^{3x}=2^{4x+6}.

    Equate the exponents.
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    Quote Originally Posted by azza View Post
    solve, giving your answers as exact fractions, the simultaneous equations:

    8^(x) = 4^(2x+3)

    log(base2)y = log(base2)x + 4
    8^x = 4^{2x+3}

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

    2^{3x} = 2^{4x+6}

    3x = 4x+6


    I assume the right side of the second equation is \log_2{x} + 4 rather than \log_2(x+4) ...

    \log_2{y} = \log_2{x} + 4

    \log_2{y} = \log_2{x} + \log_2{16}

    \log_2{y} = \log_2(16x)

    y = 16x

    can you finish it?
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  4. #4
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    sorry i typed it in incorrecty

    8^(y) = 4^(2x+3)

    log(base2)y = log(base2)x + 4


    so that gives you 3y = 4x +6 which i make into y = 4/3x + 2 and I put it into the y of the other equation. but how do i solve it then?
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  5. #5
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    y = 16x ... substitute 16x for y into the first equation.
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    thankyou very much!
    i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11
    is the textbook wrong or am i?
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  7. #7
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    Quote Originally Posted by azza View Post
    thankyou very much!
    i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11
    is the textbook wrong or am i?
    the textbook is correct ...

    3(16x) = 4x+6

    recheck your algebra.
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  8. #8
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    ahh finally i understand, thankyou so much!
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    I have ONe other question:

    log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2)

    Thankyou x
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  10. #10
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    Quote Originally Posted by azza View Post
    I have ONe other question:

    log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2)

    Thankyou x
    note that \displaystyle \log_9{u} = \frac{\log_3{u}}{\log_3{9}}

    next time start a new problem w/ a new thread.
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    sorry, i dont see where that leads?
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  12. #12
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    \log_3(2-3x) = \log_9(6x^2 - 19x + 2)

    \displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{\log_3{9}}

    \displaystyle \log_3(2-3x) = \frac{\log_3(6x^2 - 19x + 2)}{2}<br />

    \log_3(2-3x) = \log_3 {\sqrt{6x^2 - 19x + 2}}

    finish it
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  13. #13
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    Another way of looking at it: [tex]log_a(x)= y if and only if x= a^y so if y= log_3(2- 3x) then 2- 3x= 3^y[tex] and if y= log_9(6x^2- 19x+ 2) then 6x^2- 19x+ 2= 9^y.

    But 9= 3^2 so 6x^2- 19x+ 2= (3^2)^y= 3^{2y}= (3^y)^2= 2- 3x.

    Solve 6x^2- 19x+ 2= 2- 3x
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  14. #14
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    Quote Originally Posted by HallsofIvy View Post
    Another way of looking at it: [tex]log_a(x)= y if and only if x= a^y so if y= log_3(2- 3x) then 2- 3x= 3^y[tex] and if y= log_9(6x^2- 19x+ 2) then 6x^2- 19x+ 2= 9^y.

    But 9= 3^2 so 6x^2- 19x+ 2= (3^2)^y= 3^{2y}= (3^y)^2= 2- 3x.

    Solve 6x^2- 19x+ 2= 2- 3x
    the last equation should be ...

    6x^2- 19x+ 2= (2- 3x)^2
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