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    Would appreciate if someone could help:

    1. Find the possible values of x for which 2^(2x+1) = 3(2^x) - 1

    2. Solve log 3 Z = 4 log z 3

    I know the solutions 1. x=-1,0 and 2. 1/9, 9. But I don't know how to work it out.

    Rep for those who can help be sucessfully on both Questions.
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    (Original post by blah888)
    Would appreciate if someone could help:

    1. Find the possible values of x for which 2^(2x+1) = 3(2^x) - 1

    2. Solve log 3 Z = 4 log z 3

    I know the solutions 1. x=-1,0 and 2. 1/9, 9. But I don't know how to work it out.

    Rep for those who can help be sucessfully on both Questions.
    1) Let y = 2^x

    Then 2y^2 = 3y - 1 which has solutions y = 1 and y = 1/2.

    So 2^x = 1 or 1/2

    So x = 0 or -1

    2) log 3 Z = 4 log z 3

    Changing to a common base

    lnz/ln3 = 4 ln3/lnz

    (lnz)2 = 4(ln3)2

    lnz = ±2ln3 = ±ln9

    z = 9 or 1/9
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    (Original post by RichE)
    1) Let y = 2^x

    Then 2y^2 = 3y - 1 which has solutions y = 1 and y = 1/2.

    So 2^x = 1 or 1/2

    So x = 0 or -1

    2) log 3 Z = 4 log z 3

    Changing to a common base

    lnz/ln3 = 4 ln3/lnz

    (lnz)2 = 4(ln3)2

    lnz = ±2ln3 = ±ln9

    z = 9 or 1/9

    Thanks a lot RichE! Rep for you!

    But can you explain why In z = +/- In9 leads to z = 9 or 1/9

    Edit: nvm I think i get it cause -1 In 9 is the same as In (1/9) = In 9^-1 .... I'm not that dumb after all
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    (Original post by blah888)
    Perhaps you can help me with this one as well:

    The answer is: 12.

    Question:

    2 log3y - log 3(y+4) = 2

    Thanks a lot!
    Rearrange to

    log3(y²/(y+4)) = 2

    y²/(y+4) = 9

    y² - 9y - 36 = 0

    (y+3)(y-12) = 0

    y = 12 or y = -3 (but -3 as an answer makes no sense as can't take logs of negative numbers)
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    2) log 3 Z = 4 log z 3

    Changing to a common base

    lnz/ln3 = 4 ln3/lnz

    (lnz)2 = 4(ln3)2

    lnz = ±2ln3 = ±ln9

    z = 9 or 1/9[/QUOTE]

    Riche, I don't understand this.Can you please look at the picture and explain why it is OK to divide one side by In3 and the other side by Inz

    Riche, I have worked out the solution a different way. I agree with your solution.
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    (Original post by steve2005)
    2) log 3 ...
    Riche, I don't understand this.Can you please look at the picture and explain why it is OK to divide one side by In3 and the other side by Inz
    ...
    riche didn't divide one side by ln3 and t'other by lnz. He changed both sides to a common base.
    You did this yourself, in effect, in your 2nd attachment!

    you have,

    log_3 z = 4log_z 3 --------------------(1)

    change of base (lhs)

    log_3 z = log_e z/ log_e 3 = lnz/ln3

    change of base (rhs)

    log_z 3 = log_e 3 / log_e z = ln3/lnz

    applying these changes of base to the original eqn (1)

    lnz/ln3 = 4*ln3/lnz
    etc.
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    Thanks.

    I now see.
 
 
 
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