Answer:
x = 2
Step-by-step explanation:
Taking antilogs, you have ...
2³ × 8 = (4x)²
64 = 16x²
x = √(64/16) = √4
x = 2 . . . . . . . . (the negative square root is not a solution)
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You can also work more directly with the logs, if you like.
3·ln(2) +ln(2³) = 2ln(2²x) . . . . . . . . . . . write 4 and 8 as powers of 2
3·ln(2) +3·ln(2) = 2(2·ln(2) +ln(x)) . . . . use rules of logs to move exponents
6·ln(2) = 4·ln(2) +2·ln(x) . . . . . . . . . . . . simplify
2·ln(2) = 2·ln(x) . . . . . . . . . . . subtract 4ln(2)
ln(2) = ln(x) . . . . . . . . . . . . . . divide by 2
2 = x . . . . . . . . . . . . . . . . . . . take the antilogs
Go on chegg and ask your question
30 circuits is 10% of 300 circuits
I found h max = 64 feet
Explanation: Ok...probably you can do this differently but I would try to find the vertex of the parabola describing the trajectory: 1) derive it: h ` ( t ) = 64 − 32 t 2) set derivative equal to zero: 64 − 32 t = 0 t = 64 32 = 2 sec 3) use this value of t into your trajectory: h ( 2 ) = h max 64 ⋅ 2 − 16 ⋅ 4 = 64 feet .
