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solve each equation 1) log5-log2x=1 2) 2log(x+1)=5

  • math -

    1)

    log5 - log(2x) = 1
    log(5/2x) = 1
    5/(2x) = 10^1
    20x = 5
    x = 1/4

    2)

    2log(x+1) = 5
    log(x+1) = 5/2
    10^(5/2) = x+1
    316.227766 = x+1
    x = 315.227766

    OR

    log(x+1)^2 = 5
    (x+1)^2 = 10^5
    x+1 = ±√(10^5) , but clearly x >-1
    x+1 = 316.227766..
    x = 315.227766..

  • math -

    1)> log 5 + log 2 + log x = 1
    >log (5 X 2) + log x = 1
    >1 + log x = 1
    >log x = 0
    >x = 10^0
    >x = 1

    #2 has me twisted.

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