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Pre-Cal Help Please!!

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Use logarithms to solve the given equation. (Round the answer to four decimal places.)
6(1.02^3x + 1) = 11

How do i find the answer after
1.03^3x +1 = (11/6)
Do i put a log on 1.03 or do I do something else

Thank you.

  • Pre-Cal Help Please!! -

    1.03^3x +1 = (11/6)
    the way you typed it:
    1.03^3x = 11/6 - 6/6 = 5/6

    3x (log 1.03) = log 5 - log 6
    3 x = 48.72
    x = 16.24
    if you meant to type
    1.03^(3x +1) = (11/6)
    then
    (3x+1)log1.03 = log 11 - log 6
    3x+1 = 78.64
    x = 25.9

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