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Math Help - Logarithms

  1. #1
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    Logarithms

    I understand how to do problems such as (2)(3^x)=(7)(5^x) but I'm stuck on these problems:

    a) (4.6)(1.06^(2x+3))=(5)(3^x)

    b) (2.67)(7.38^x)=(9.36^(5x-2))

    c) (7)(0.43^(2x))=(9)(6^-x)

    Any help would be greatly appreciated. Thanks.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by andymac View Post
    I understand how to do problems such as (2)(3^x)=(7)(5^x) but I'm stuck on these problems:

    a) (4.6)(1.06^(2x+3))=(5)(3^x)
    You do them in the same way.
    ln \left ( 4.6 \cdot 1.06^{2x + 3} \right ) = ln \left ( 5 \cdot 3^x \right )
    (or you can take log_a( ) of both sides where a is your favorite base.)

    ln(4.6) + (2x + 3) \cdot ln(1.06) = ln(5) + x \cdot ln(3)
    which is just a linear equation for x. Solve for x.

    -Dan
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  3. #3
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    Sorry, I'm still not getting the right answer when I solve for x. Could you show me the steps you would take in order to do so? Thanks, and sorry about the double post.
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by topsquark View Post
    You do them in the same way.
    ln \left ( 4.6 \cdot 1.06^{2x + 3} \right ) = ln \left ( 5 \cdot 3^x \right )
    (or you can take log_a( ) of both sides where a is your favorite base.)

    ln(4.6) + (2x + 3) \cdot ln(1.06) = ln(5) + x \cdot ln(3)
    which is just a linear equation for x. Solve for x.

    -Dan
    Taking it up from where I left off:
    2~ln(1.06) \cdot x + 3 \cdot ln(1.06) + ln(4.6) = ln(3) \cdot x + ln(5)

    (2~ln(1.06) - ln(3))x = ln(5) - 3 \cdot ln(1.06) - ln(4.6)

    x = \frac{ln(5) - 3 \cdot ln(1.06) - ln(4.6)}{2~ln(1.06) - ln(3)} \approx 0.093094

    -Dan
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  5. #5
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    Smile

    Alright thanks a lot. I get it now.
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