Please help me!!!! maths

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Solve for x: log5(x-2)+log8(x-4) = log6(x-1). I do not have any idea on this topic.

  • Please help me!!!! maths -

    I am not sure if the 5 , 8 , and 6 are bases of the logs
    or just multipliers.
    Let's hope they are just multipliers, or else it would be a terrible terrible mess

    by the laws of logs
    log5(x-2)+log8(x-4) = log6(x-1)
    log [40(x-2)(x-4)] = log 6(x-1)
    anti-log it

    40(x^2 - 6x + 8) = 6x-6
    40x^2 - 240x + 320 - 6x + 6 = 0
    40x^2 - 246x + 326 = 0
    20x^2 - 123x + 163 = 0
    by the formula
    x = (123 ± √2089)/40
    = 4.2176 or 1.9324 , but from the original equation , x > 4

    so x = 4.2176

    I tested the answer, it works

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