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Solve for x in the following equations. (4 marks)

log3(x - 5) + log3(x + 3) = 2 <--- both log3.. 3 is subscript

log7x + log7(x - 1) = log72x <--- all log7.. 7 is subscript.

  • math -

    use your log rules:
    log (AB) = logA + logB
    and
    log (A/B) = logA - logB

    1st one:
    log3 [(x-5)(x+3)] = 2
    by definition of logs
    (x-5)(x+3) = 3^2 = 9
    x^2 - 2x -24=0
    (x-6)(x+4) = 0
    x = 6 or x=-4 , but x=-4 would make log3(x+3) undefined, so

    x = 6

    2nd:
    log7(x) + log7(x-1) - log7(2x) = 0

    log7[x(x-1)/(2x)] = 0
    log7[(x-1)/2] = 0
    (x-1)/2 = 7^0
    (x-1)/2 = 1
    x-1=2
    x = 3

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