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    :stupido2:

    How would i find the values of x for which

    Log3 X – 2logx 3 = 1
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    The rule for converting unnatural logs into natural ones is

    log[a](b) = ln(b)/ln(a)

    Using that rule twice,

    ln(x)/ln(3) - 2ln(3)/ln(x) = 1
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0
    (ln(x) + ln(3))(ln(x) - 2ln(3)) = 0
    ln(x) = -ln(3) or 2ln(3)
    ln(x) = ln(1/3) or ln(9)
    x = 1/3 or 9
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    (Original post by Jonny W)
    The rule for converting unnatural logs into natural ones is

    log[a](b) = ln(b)/ln(a)

    Using that rule twice,

    ln(x)/ln(3) - 2ln(3)/ln(x) = 1
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0
    (ln(x) + ln(3))(ln(x) - 2ln(3)) = 0
    ln(x) = -ln(3) or 2ln(3)
    ln(x) = ln(1/3) or ln(9)
    x = 1/3 or 9
    Excuse my stupidity but im lost after
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0 and the rest of the lines
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    (Original post by ThugzMansion7)
    Excuse my stupidity but im lost after
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0 and the rest of the lines
    Let y = ln(x) and k = ln(3). Then the equation is

    y^2 - y*k - 2k^2 = 0

    which factorises to

    (y + k)(y - 2k) = 0

    Now replace y by ln(x) and k by ln(3).
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    (Original post by ThugzMansion7)
    Excuse my stupidity but im lost after
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0 and the rest of the lines
    any1?
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    log_3_x–2log_x_3 = 1

    By using change of base: log_b_a=(lna/lnb)

    Now can you see how log_3_x=(lnx/ln3) and log_x_3=(ln3/lnx)

    Therefore, if we let log_3_x=a then log_x_3=(1/a)

    Hence, by substitution:

    a-(2/a)=1

    Multiply through by to give:

    (a^2)-2=a

    Therefore:

    (a^2)-a-2=0

    =>(a+1)(a-2)=0

    Hence a=-1 OR a=2 => log_3_x=-1 or log_3_x=2

    Hence from first principles:

    x=(1/3) OR x=9

    Newton.
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    ln(x)/ln(3) - 2ln(3)/ln(x) = 1
    ln(x)^2 - ln(x)ln(3) - 2ln(3)^2 = 0

    how do you get from the first line to the second. thats the only thing i find a little confusing.

    the rest is okay
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    You want to get a common denominator so you multiply through by ln3lnx. Same answer, but I prefer my way... Looks neater!

    Newton.
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    (Original post by Newton)
    You want to get a common denominator so you multiply through by ln3lnx. Same answer, but I prefer my way... Looks neater!

    Newton.
    Multiply through by to give:

    (a^2)-2=a

    how'd u do that?
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    It's just like getting a common denominator:

    a-(2/a)=1

    From simple algebra, "What you do to one side you do to the other!". Right? So...

    a(a-(2/a))=a(1)

    Which gives:

    a(a)-a(2/a)=a(1)

    So:

    (a^2)-(the as cancel)2=a

    And then rearrange to give:

    (a^2)-a-2=0

    Newton.
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    (Original post by Newton)
    It's just like getting a common denominator:

    a-(2/a)=1

    From simple algebra, "What you do to one side you do to the other!". Right? So...

    a(a-(2/a))=a(1)

    Which gives:

    a(a)-a(2/a)=a(1)

    So:

    (a^2)-(the as cancel)2=a

    And then rearrange to give:

    (a^2)-a-2=0

    Newton.

    LoL...Sorry its just that simple eq's and stuff are quite hard to see as they're typed up...it makes me ask funny questions!
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    Do you understand it now? It's basically like getting a common denominator:

    a-(2/a)=1

    You only have a fraction on the left side so get all of it into a fraction:

    (a/1)-(2/a)=1

    The L. C. M. on the L. H. S. is a so you multiply the top and bottom of (a/1):

    ((a*a)/(1*a)-(2/a)=1

    Therefore:

    (((a^2)-2)/a)=1

    You can think of it in many different ways. One is sort of to bring the denominator from the L. H. S. over to the right to leave you without fractions to work with... So:

    (a^2)-2=a

    Therefore:

    (a^2)-a-2=0

    But can you see thinking of it as "MULTIPLYING THROUGH" actually is the same thing, but it gets you there quicker as there is less mental thinking going on!

    Capish?

    Newton.
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    Ok... Posted the last one before I saw your answer. Anyway feel free to ask! (So can I have some rep!?) . Lol.

    Newton.
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    (Original post by Newton)
    Ok... Posted the last one before I saw your answer. Anyway feel free to ask! (So can I have some rep!?) . Lol.

    Newton.
    u soooo fully deserve it...thing is i gave it to gaz earlier on 2day. I will send u some 2mrw if im allowed...

    one more question...what did you mean by "hence from first principles"

    how did u get the answer by saying that log3X = -1


    P.S TSR's been a bit wierd, when i refresh the page some post's will either be missing or they won't appear until i've posted again..like what happened to u
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    Well...

    The logarithm of b to base a is basically to what power we need to raise a to get b, let us say this is x. This is shown by log_a_b=x

    So this means that (a^x)=b (One of the first principles of logarithms)

    Just like in your case where it says log_3_x=-1. This means that the logarithm of x to base 3 is -1 OR we need to raise 3 to -1 to give x. This means:

    3^(-1)=x

    Since: z^(-n)=(1/(z^n))

    Therefore:

    x=1/3

    And similarly:

    log_3_x=2 (Say it with me): The logarithm of x to base 3 is 2, and hence we need to raise 3 to 2 to give x

    Therefore:

    x=(3^2)=9

    Newton.
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    (Original post by Newton)
    Well...

    The logarithm of b to base a is basically to what power we need to raise a to get b, let us say this is x. This is shown by log_a_b=x

    So this means that (a^x)=b (One of the first principles of logarithms)

    Just like in your case where it says log_3_x=-1. This means that the logarithm of x to base 3 is -1 OR we need to raise 3 to -1 to give x. This means:

    3^(-1)=x

    Since: z^(-n)=(1/(z^n))

    Therefore:

    x=1/3

    And similarly:

    log_3_x=2 (Say it with me): The logarithm of x to base 3 is 2, and hence we need to raise 3 to 2 to give x

    Therefore:

    x=(3^2)=9

    Newton.


    DOH! thanks ever so much for ur help

    24hours till i can rep agen
 
 
 
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