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    see the pic, I don't know how to 'show it' except solving for x (1..9) then adding them all together.
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    Isnt it an arithmetic progression? so u can solve it that way?
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    U_2 - U_1 != U_3 - U_2
    U_2/u_1 != U_3/u_2
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    Write out the terms and use one of the laws of logarithms [loga+logb=logab] to show that the terms are equal to log(10)10=1.
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    (Original post by Revelation)
    see the pic, I don't know how to 'show it' except solving for x (1..9) then adding them all together.
    You have to prove that the bits inside the brackets multiply to give 10.

    ==> log(2) + log(3/2) + log(4/3) + log(5/4) + log(6/5) + log(7/6) + log(8/7) + log(9/8) + log(10/9)

    ==> log( 10! / 9! )

    ==> log(10) ==> 1

    It is a bit of a crap question if you ask me.
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    mm, it has something to do with "Benford's Law" which is breifly explained with the question but I still dont get it. I understand how to do the question tho, thanks
 
 
 
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Updated: June 13, 2005
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