PreCal Help Please!!
posted by Ann .
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.

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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