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Answer:
4) -ln(2)/3
5) x = 9
6) x = 2
7) x = -6
Step-by-step explanation:
4) e^(-3x) = 2 . . . . given
-3x = ln(2) . . . . . take the natural log
x = -ln(2)/3 . . . . divide by the coefficient of x
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5) ln(3x -5) = ln(11) +ln(2)
3x -5 = 11·2 . . . . . take antilogs
3x = 27 . . . . . . . . add 5
x = 9 . . . . . . . . . . divide by 3
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6) log4(x -1) +log4(x +2) = 1
(x -1)(x +2) = 4 . . . . . take antilogs
x^2 +x -2 = 4 . . . . . . eliminate parentheses
x^2 +x +1/4 = 6 1/4 . . . add 2 1/4 to complete the square
(x +0.5)^2 = 6.25 . . . . write as a square
x +0.5 = ±2.5 . . . . . . . take the square root
x = -0.5 ± 2.5 = {-3, +2} . . . . . . x = -3 is an extraneous solution
x = 2
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7) 8^x = 4^(x -3)
2^(3x) = 2^(2(x -3)) . . . . write using a common base of 2
3x = 2(x -3) . . . . . . . . . . take logarithms base 2
x = -6 . . . . . . . . . . . . . . .subtract 2x
