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    8). Solve the equation log_3(2-3x)=log_9(6x^2-19x+2)

    I changed the base and got to: 2log_9(2-3x)=log_9(6x^2-19x+2)

    I tried to use the power rule remove the coefficient of 2 on the LHS and raise the log on the LHS to 2, but according to wolfram that gives a different answer.

    Without Power Rule
    With Power Rule
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    At your stage, you really oughtn't be using Wolfram. However, you've inputted the expression incorrectly (the way you've put it Wolfram thinks you mean log * log = log^2, rather than log(whatever^2):

    http://www.wolframalpha.com/input/?i...5E2-19x%2B2%29
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    Ahh ok thanks. What should I be trying to do from this point though? Am I able to solve the quadratic, or expand the (2-3x)^2 on the LHS? Or do the logs stop me from doing that?
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    I got a question im stuck on atm.

    4(3^2x+1) + 17(3x) - 7.


    I let y = 3x. But somehow got muddled up with the first part of the substition.

    Any help appreciated.
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    hmmm
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    (Original post by ViralRiver)
    8). Solve the equation log_3(2-3x)=log_9(6x^2-19x+2)

    I changed the base and got to: 2log_9(2-3x)=log_9(6x^2-19x+2)

    I tried to use the power rule remove the coefficient of 2 on the LHS and raise the log on the LHS to 2, but according to wolfram that gives a different answer.

    Without Power Rule
    With Power Rule

    Change the log_9() base to a base of log 3()
    so you get  log_3(6x^2-19x+2) /  log_3(9)
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    (Original post by Demonic Entity)
    I got a question im stuck on atm.

    4(3^2x+1) + 17(3x) - 7.


    I let y = 3x. But somehow got muddled up with the first part of the substition.

    Any help appreciated.
    yeah
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    (Original post by Demonic Entity)
    yeah
    Get your own thread and stop derailing someone else's.
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    If you have log(a) = log(b), you can conclude that a = b (if they're in the same base).

    i.e. (2-3x)^2 = 6x^2 - 19x + 2.

    See, these questions aren't so hard, are they?
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    Oh, that makes more sense. So is this true:

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

    EDIT: Hmm, I see you used the power rule, does that mean that that if log_a(b)^c = log_a(d)^e then b^c=d^e even though the indices are different?
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    I'm stuck on the mixed exercise as well >_<
    Please may someone kindly help?

    question no. 7
    Find the values of x for which log_3x-2log_x3=1

    this is as far I have got into in number 7, my workings is as follows:

    log_3x-2log_x3=1
    log_3x+2/log_x3=1
    let log_3x=y
    y+2/y=1
    y^2+2=y
    y^2-y+2=0
    (y-1)(y+2)=0 or (y-1)(y-2)=0?? <--- this is the bit I'm stuck on, you gotta factorise here, but either way (y-1) & (y+2) together or (y-1) & (y-2) together added together won't give -y, it will only give 2 or -3.

    please may someone help??

    --------------------------

    I am also stuck on question 12.

    a) Given that 3+2log_2x=log_2y, show that y=8x^2.
    b) Hence, or otherwise, find the roots α and β, where α < β, of the equation 3+2log_2x=log_2(14x-3).
    c)Show that log_2α=-2
    d) Calculate log_2β,giving youranswer to 3 significant figures.

    Help may someone help me, your help will be greatly appreciated.
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    I'd appreciate it if you kept your questions in a separate topic . Although I did complete question 7:

    2log_x(3)=\frac{2}{log_3(x)}
    (log_3(x))^2-2=log_3(x)
    (log_3(x))^2=log_3(x)+2
    y=log_3(x)\rightarrow y^2=y+2
    y^2-y-2=0
    (y+1)(y-2)=0
    y=-1 or y=2

    y=-1\rightarrow log_3(x)=-1
    3^{-1}=x
    x=\frac{1}{3}

    y=2\rightarrow log_3(x)=2
    x=3^2=9
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    (Original post by ViralRiver)
    Oh, that makes more sense. So is this true:

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

    EDIT: Hmm, I see you used the power rule, does that mean that that if log_a(b)^c = log_a(d)^e then b^c=d^e even though the indices are different?
    Your notation here is bad. If you mean

    \log_a(b^c) = \log_a(d^e)

    Then yes, you can conclude that b^c = d^e. Why? Just let f = b^c and let g = d^e, then

    \log_a(b^c) = \log_a(d^e)
    \log_a(f) = \log_a(g)
    f = g
    b^c = d^e
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    Sorry, what I meant was, if (log_a(b))^c=(log_a(d))^e then b^c=d^e?
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    (Original post by ViralRiver)
    Sorry, what I meant was, if (log_a(b))^c=(log_a(d))^e then b^c=d^e?
    No, because (log_a(b))^c\neq log_a(b^c)
 
 
 
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