advanced algebra

we are working with logarithms and i don't understand them at all! I get the basic content but then it gets a little more advanced, and I get completely confused.

There's one in particular that I have NO idea how to do. Help please?!

Simplify:
3 ^ [log(base 9)2 + log(base 9)4]

Thank you.. =]

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asked by Kate
  1. when you add logs of the same base, you are multiplying
    like
    log A + log B = log (A*B)
    so we have really

    3^log9(8)
    --------------------------
    ONE WAY --If you know log rules
    logb(M) = logc(M)/logc(b)
    so
    log9(8) = log3(8)/log3(9)
    but
    log3 (9) = log3 (3)^2 = 2*1 = 2
    so
    log9(8) = log 3(8)/2 = (1/2)log3(8)
    so now back to the problem
    3^log9(8) = 3^(1/2) log3(8)
    = (3^log3(8))^1/2
    but b^logb(x) = x
    so this is 8^(1/2)
    = sqrt(8) = 2 sqrt(2)
    ----------------------------
    Another more basic way:
    Now a basic fact of logs is that base^log x = x
    so 9^log9(8) = 8

    so (3^2)^log9(8) = 8

    so 3^2log9(8) = 8

    so [3^log9(8)]^2 = 8

    so 3^log9(8) = sqrt 8 = 2 sqrt 2

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    posted by Damon

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