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Math Help - Solving equations involving logs

  1. #1
    Junior Member Fnus's Avatar
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    Solving equations involving logs

    Help with these two would be grealy appreciated.
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  2. #2
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    Quote Originally Posted by Fnus View Post
    Help with these two would be grealy appreciated.
    2. From equation (2): 2x = 9 - 3y \Rightarrow x = \frac{9-3y}{2}.

    Substitute this into equation (1):

    2^{\frac{9-3y}{2}} \cdot 2^y = 16 \Rightarrow 2^{\frac{9-3y}{2} +y} = 2^4 \Rightarrow \frac{9-3y}{2} +y = 4 .......
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  3. #3
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    Quote Originally Posted by Fnus View Post
    Help with these two would be grealy appreciated.
    1. On the left hand side, you're missing an argument for the log. Nevertheless:

    Note that 2 = \log_5 25 and so the right hand side is \log_5 25 + \log_5 2 = \log_5 50.

    The left hand side can also be expressed as a single log. Do it.

    Now equate the arguments of the logs on each side and solve for x.
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  4. #4
    Junior Member Fnus's Avatar
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    Yay, I get y = 1, and x = 3, and I'm pretty sure that checks out.
    Thanks a ton. (:
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  5. #5
    Junior Member Fnus's Avatar
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    Oh, yeah, on the left side I'm missing an x.

    But I don't get what it is you're trying to do by <br />
\log_5 25 + \log_5 2 = \log_5 50<br />
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  6. #6
    Flow Master
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    Quote Originally Posted by Fnus View Post
    Oh, yeah, on the left side I'm missing an x.

    But I don't get what it is you're trying to do by <br />
\log_5 25 + \log_5 2 = \log_5 50<br />
    Then the left hand side can be written \log_5 [(x-5) x].

    The right hand side can be written \log_5 50.

    Therefore (x - 5) x = 50 \Rightarrow x^2 - 5x - 50 = 0 \Rightarrow x = .....

    Warning: Although the quadratic equation has two solutions, only one is valid (why?).
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  7. #7
    Junior Member Fnus's Avatar
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    Oh *okay*.

    And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..
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  8. #8
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    Quote Originally Posted by Fnus View Post
    Oh *okay*.

    And I get x = -5 or 10, but it can't be -5 because you can't take the logarith of a negative number, I think..
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  9. #9
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    Another way for the 2nd one

    2^x \cdot 2^y=16 \iff 2^{x+y}=2^4

    Now we note from here the bases are the same so the exponents must be equal. So we get x+y=4 now we multiply it by -2 and add to the one below to get

    2x+3y=9
    -2x-2y=-8
    y=1

    Subbing back in we get
    x=3
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