Question 75640
Given:
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{{{v=176*(1-0.834^t)}}}
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in which v represents velocity in ft/sec of a falling skydiver and t represents the time
in seconds since the skydiver left the plane.
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Problem.  How long does it the skydiver to reach 90% (or 0.9) of terminal velocity?
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So the velocity of the skydiver is to reach 0.9*176 ft/sec. Plug that into the equation
as the velocity and you get:
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{{{0.9*176 = 176*(1 - 0.834^t)}}}
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Note that 176 is a factor on both sides of this equation.  Therefore, if you divide both
sides by 176 you simplify the equation to:
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{{{0.9 = (1 - 0.834^t)}}}
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and the parentheses can be removed without affecting the equation:
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{{{0.9 = 1 - 0.834^t}}}
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Eliminate the 1 from the right side of this equation by subtracting 1 from both sides to
get:
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{{{ -0.1 = -0.834^t}}}
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Finally, multiply both sides by -1 to get rid of the minus signs and then transpose
(make the left side the right side and the right side the left side).  After you do those two
steps you are left with:
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{{{0.834^t = 0.1}}}
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Next you take the logarithm of both sides.  (For convenience, use the log to the base 10
because it's one of the two logarithms on most calculators.) When you do that you get:
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{{{log(0.834)^t = log(0.1)}}}
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Recall from the rules of logarithms that if you take the log of a quantity that has an
exponent, you can just multiply the exponent times the log of the quantity.  In this case
the quantity is 0.834 and the exponent is t.  Applying this rule to your equation 
converts it to:
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{{{t*log(0.834) = log(0.1)}}}
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I guess that now is as good a time as any to grab the calculator and find that log(0.834)
is equal to -0.078833949 and log(0.1) is equal to -1.  Substitute these two values into
the equation to get:
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{{{t*(-0.078833949) = -1}}}
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Now divide both sides of this equation by -0.078833949 and you find the answer to be:
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{{{t = (-1)/(-0.078833949) = 12.68489031}}} 
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So 90% of terminal velocity is reached in about 12.68 seconds ... and that's the answer you
need.
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Hope this helps you to understand that logs can be a big help in finding unknown exponents.  
Whenever you see a variable in an exponent, it may be a clue to thinking about using the
log function.
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