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    2log base 3 (x) - log base3 (x-2) = 2
    where x > 2
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    (Original post by beth59411)
    2log base 3 (x) - log base3 (x-2) = 2
    where x > 2
    Start by using some simple log laws to get the two logarithms together
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    (Original post by AHappyStudent)
    Start by using some simple log laws to get the two logarithms together
    so would it be log base 3 (x^2 /x-2) = 2?
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    (Original post by beth59411)
    so would it be log base 3 (x^2 /x-2) = 2?
    Yep, good work

    Now try using the definition of a log to find an equation for x and then solve

    By this I mean

    Where y = x^n,
    n = log base x (y)
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    (Original post by Unknown-99)
    Here's the worked solution. I wrote it down but couldn't upload the photo so I found a link with the solution.
    https://www.symbolab.com/solver/loga...2%5Cright)%3D2
    I managed to work it out thank youuu!!!!

    I got the right answer thanks for the help, I always struggle starting it off :/
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    Here is the solution I wrote it down myself but couldn't upload the photo so here's the worked solution.
    https://www.symbolab.com/solver/loga...2%5Cright)%3D2
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    (Original post by beth59411)
    I managed to work it out thank youuu!!!!

    I got the right answer thanks for the help, I always struggle starting it off :/
    No problem. Just remember these 2 simple rules.

    - If the logs are ADDED together then combine them by multiplying the things in the brackets by each other.

    - If the logs are SUBTRACTED then combine them by dividing the first bracket by the second bracket.

    But you can only do this if they have the same base.
 
 
 
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