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algebra 2 honors

posted by on .

how do you solve problems like :
log3x-log5=1

  • algebra 2 honors - ,

    log3x-log5=1
    log (3x/5) = 1
    then 3x/5 = 10^1

    3x = 50
    x = 50/3

    check:
    log(3(50/3)) - log5
    = log50-log5
    = 1.69897 - .69897
    = 1

  • algebra 2 honors - ,

    how do u solve
    (6/4x^2)+ (2/5x)

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