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Math Help - logs

  1. #1
    Junior Member
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    logs

    use the properties of logarithims to simplify the logarithmic expression:

    -log base 5 of (1/15)

    -log base 2 of (4^2 * 3^4)

    -log base of log base 5 of (1/250)

    thanks!!
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  2. #2
    Super Member

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    Hello, ss103!

    I'm not sure what they mean by "simplify", but here goes . . .



    Use the properties of logarithims to simplify the logarithmic expression:

    1)\;\;\log_5\left(\tfrac{1}{15}\right)

    \log_5\left(\frac{1}{15}\right) \;=\;\log_5(1) - \log_5(15)

    . . . . . . . = \;0 - \log_5(3\cdot5)<br />

    . . . . . . . = \;-\bigg[\log_5(3) + \log_5(5)\bigg]

    . . . . . . . = \;-\log_5(3) - \log_5(5)

    . . . . . . . = \;-\log_5(3) - 1




    (2)\;\;\log_2\left(4^2\cdot3^4\right)

    \log_2\left(4^2\cdot3^4\right) \;=\;\log_2(4^2) + \log_2(3^4)

    . . . . . . . . = \;2\log_2(4) + 4\log_2(3)

    . . . . . . . . = \;2\log_2(2^2) + 4\log_2(3)

    . . . . . . . . = \;2\!\cdot2\log_2(2) + 4\log_2(3)

    . . . . . . . . = \;4 + 4\log_2(3)




    This one was not written clearly, but I'll take a guess.

    (3)\;\;\log_5\left(\tfrac{1}{250}\right)

    \log_5\left(\frac{1}{250}\right) \;=\;\log_5(1) - \log_5(250)

    . . . . . . . = \;0 - \log_5(2\cdot5^3)

    . . . . . . . = \;-\bigg[\log_5(2) + \log_5(5^3)\bigg]

    . . . . . . . = \;-\log_5(2) - 3\log_5(5)

    . . . . . . . = \;-\log_5(2) - 3

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