2011 04 19 202231 logarithms

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2011 04 19 202231 logarithms Powered By Docstoc
					1)
a. e.05t = 1600
Answer: t = 147.556
Work:
e^(05t) = 1600
or, 0.05t = ln(1600)
or, 0.05t = 7.3778
or, t = 7.3778/0.5
or, t = 147.556


b.
ln(4x) =3
Answer: x = 5.0214
Work:
ln(4x) =3
or, 4x = e^3
or, x = e^3/4
or, x = 5.0214

c. log2(8 – 6x) = 5
Answer: x = -4
Work:
log2(8-6x) = 5
or, 8-6x = 2^5
or, 8-6x = 32
or, 6x = 8-32
or, x = -24/6
or, x = -4


d. 4 + 5e-x = 0
Answer: No solution
or, 5e^-x = -4
or, e^-x = -4/5
Here left side is positive for all values of x, while right is negative. So, no solution exists.
2) Describe the transformations on the following graph of)log()(xxf=. State the placement of the
vertical asymptote and x-intercept after the transformation. For example, vertical shift up 2 or
reflected about the x-axis are descriptions.
XY-10-9-8-7-6-5-4-3-2-112345678910-10-9-8-7-6-5-4-3-2-1123456789100

a) g(x) = log( x + 5)
Description of transformation: Left shift 5 and take log
Equation(s) for the Vertical Asymptote(s): x = -5
x-intercept in (x, y) form: (-4, 0)


b) ) g(x)=log(-x)
Description of transformation: Reflect about y-axis and take log
Equation(s) for the Vertical Asymptote(s): x = 0
x-intercept in (x, y) form: (-1, 0)




3. Students in an English class took a final exam. They took equivalent forms of the exam at
monthly intervals thereafter. The average score S(t), in percent, after t months was found to be
given by
S(t) = 68 - 20 log (t + 1), t = 0.
What was the average score when they initially took the test, t = 0?
Round your answer to a whole percent, if necessary.
a)
Answer: 68%
Show your work in this space:
S(0) = 68-20*log(0+1) = 68-20*0 = 68

What was the average score after 4 months? after 24 months?
Round your answers to two decimal places.
b)
Answer: After 4 months: 54.02%, After 24 months: 40.04%
Show your work in this space:
S(4) = 68-20*log(4+1) = 54.02
S(24) = 68-20*log(24+1) = 40.04
After what time t was the average score 50%?
Round your answers to two decimal places.
c)
Answer: t = 6.94
Show your work in this space:
50 = 68 - 20 log (t + 1)
Or, 20log(t+1) = 68-50
Or, log(t+1) = 18/20
Or, t+1 = 10^(18/20) = 7.9433
Or, t = 7.9433-1 = 6.94




4) The formula for calculating the amount of money returned for an initial deposit into a bank
account or CD (certificate of deposit) is given by ntnrPA??????+=1 A is the amount of the
return. P is the principal amount initially deposited. r is the annual interest rate (expressed as a
decimal). n is the number of compound periods in one year. t is the number of years.
Carry all calculations to six decimals on each intermediate step, then round the final answer to
the nearest cent. Suppose you deposit $2,000 for 5 years at a rate of 8%.

a) Calculate the return (A) if the bank compounds annually (n = 1). Round your answer to the
nearest cent.
Answer: A = $2,938.66
Show work in this space. Use ^ to indicate the power or use the Equation Editor in MS Word.
A = $2000*(1+0.08/1)^(1*5) = $2,938.66



b) Calculate the return (A) if the bank compounds quarterly (n = 4). Round your answer to the
nearest cent.
Answer: A = $2,971.89
Show work in this space:
A = $2000*(1+0.08/4)^(4*5)= $2,971.89



c) If a bank compounds continuously, then the formula used is rtPeA= where e is a constant and
equals approximately 2.7183. Calculate A with continuous compounding. Round your answer to
the nearest cent.
Answer: A = $2,983.66
Show work in this space: A = $2000*(2.7183)^(0.08*5) = $2,983.66

				
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