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Math Help - 2 problems i need help on.

  1. #1
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    2 problems i need help on.

    couple problems i would appreciate getting help on.

    9 ^ x2 = 3 ^ x/5 solve for x.

    the up arrow means the following is a root.


    also...

    log(x^2 + 3x + 2) - 2log(x+1) ..... write as a single logarithm.



    any help on either or both would be real tight.
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  2. #2
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    Solution to log problem

    log(x^2 + 3x + 2) - 2log(x+1) ..... write as a single logarithm.

    Okay, so the first place to start is to rewrite the term 2log(x+1) as log(x+1)^2 because log contains the property logx^n = n*logx. So the expression now looks like:

    log(x^2 + 3x + 2) - log(x+1)^2.

    Another property of log is that logx - logy = log(x/y). So the above can be rewritten as:

    log((x^2 + 3x +2)/(x+1)^2)

    But x^2 +3 + 2 can be factored into (x+1)(x+2). So you can cancel an (x+1) term from both the numerator and denominator. The final expression should look like:

    log((x+2)/(x+1)).

    If I'm not mistaken I don't think it can be simplified anymore. I hope this is useful.
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  3. #3
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    Possible solution to problem 1

    9 ^ x2 = 3 ^ x/5 solve for x.

    I have one question. Is 9^x2 read as 9 raised to the x squared power. Because if it is then I may have a possible solution:

    9^(x^2) = 3^x/5

    9^(x^2) can be rewritten as 3^(2*x^2) Since 9 = 3^2.

    Now take the log of both sides:

    2x^2log3 = (x/5)log3

    Cancel the log 3 terms from both sides to get:

    2x^2 = x/5.

    subtract x/5 from both sides to get:

    2x^2 - x/5 = 0

    Factor out an x from each term to yield:

    x(2x - 1/5) = 0.

    Set each equal to zero. You get the solutions:

    x = 0, 2x - 1/5 = 0
    2x = 1/5
    x = 1/10

    x=0, x = 1/10

    Perhaps you could double check with a classmate to confirm this.
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  4. #4
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    thnx buddy
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