Transcript Math 140 Q4 Solutions - City Colleges of Chicago

Math 140
Quiz 4 - Summer 2004
Solution Review
(Small white numbers next to problem number
represent its difficulty as per cent getting it wrong.)
Problem 1
(33)
Approximate each value using a calculator.
Express answer rounded to three decimal places.
log9(1/81)
Without a calculator one notes,
log9(1/81) = log9(9-2) = -2log9(9) = -2.
Or, evaluate one, or other, change-of-base formula:
log9(1/81) = log(1/81)/log(9),
log9(1/81) = ln(1/81)/ln(9).
Problem 2
(06)
Approximate each value using a calculator.
Express answer rounded to three decimal places.
e2.64
If necessary, use keystrokes 1, INV(or 2nd), ln to get
value of e.
Then, press yx key and enter 2.64 followed by equal
(or enter) key.
Result is 14.01320361..., which rounds to 14.013.
Problem 3
Convert to logarithmic form:
(28)
53 = 125.
One takes the base-5 logarithm of both sides to get
log5(53) = log5(125).
Then one observes that log5(53) = 3 log5(5) = 3.
Thus,
log5(125) = 3.
Problem 4
(06)
Use a calculator to find the natural logarithm
correct to four decimal places.
ln(821/2)
One notes,
ln(821/2) = (1/2)ln(82) = 2.203359624 = 2.2034.
Or, evaluate the square root first and find:
ln(821/2) = ln(9.055385138) =
2.203359624 = 2.2034.
Problem 5
(50)
If the following defines a one-to-one function, find
the inverse of:
f = {(-3, -7), (-2, -7), (-1, -1), (0, 2)}.
Since both domain values –3 and –2 lead to –7,
f fails the horizontal line test and is not 1-to-1.
If one proceeded by swapping x <=> y in f , getting:
f -1 = {(-7, -3), (-7, -2), (-1, -1), (2, 0)}? But this is
not a function and would not check. For example,
f -1(f(-3)) = f -1(-7) = -3 or -2 and not just -3.
Problem 6
(11)
If the following defines a one-to-one function, find
the inverse of:
f(x) = 2x + 8.
The horizontal line test is satisfied since a graph of
_ y = f(x) = 2x + 8 is a line. Hence, f(x) is 1-to-1.
One proceeds by swapping x <=> y and solving
for y. Thus,
x = 2y + 8
=>
y = (x - 8)/2.
This means that f -1(x) = (x - 8)/2.
Check: f -1(f(x)) = [(2x + 8) - 8]/2 = x.
Problem 7
(61)
Compute the amount in m years if a principal P is
invested at a nominal annual interest rate of r
compounded as given. Round to the nearest cent.
P = $480, m = 7, r = 7% compounded quarterly
A = P (1 + r/n )nm
= $480 (1 + 0.07/4 )(4)(7)
= $480 (1.0175 )28
= $780.20
Problem 8
(56)
Graph f(x) as a solid line/curve & its inverse as a
dashed line/curve on the same axes. f(x) = 4x.
Recognize f(x) as a straight line
of slope m = 4 and y-intercept
(0,0).
Draw the line L: y = x.
Construct the inverse function
line by reflection in L.
Or, solve x = 4y for y to get
L
f- -1(x) = x/4.
Then graph it.
This is answer C).
Problem 9
(61)
Use a calculator and the base conversion formula
to find the logarithm, correct to three decimal
places.
log6.9(5.5)
Evaluate one, or other, change-of-base formula:
log6.9(5.5) = log(5.5)/log(6.9) =
0.74036/0.83885 = 0.883,
log6.9(5.5) = ln(5.5)/ln(6.9) =
1.70475/1.93152 = 0.883.
Problem 10
(72)
Decide whether or not f(x) and g(x) are
inverses of each other with:
a) f(x) = 3x + 5, g(x) = (1/3)x - 5;
b) f(x) = (x - 6)2, g(x) = x1/2 + 6.
One needs to see if:
(g o f )(x) = g(f(x)) = x or (f o g )(x) = f(g(x)) = x.
a) f(g(x) ) = 3[(1/3)x - 5] + 5 = x - 10 => NO.
b) f(g(x)) = [(x1/2 + 6) - 6]2 = [x1/2]2 = x => YES.
Problem 11
(83)
How long will it take for an investment to triple in
value if it earns 8.75% compounded continuously?
Round your answer to three decimal places.
A = Pert
3P = Pe0.0875t
3 = e0.0875t
ln(3) = ln(e0.0875t ) = 0.0875t ln(e) = 0.0875t
0.0875t = ln(3) =>
t = ln(3)/0.0875 = 12.556 yr
Problem 12
(39)
Select the matching logarithmic function to the graph.
Eliminate A), B), C), and E),
A) log(1/3)
B)
C)
E)
log(1)
3log(3
– log(1)
––1)
=
3=
– .48
–30.3
3
choices by evaluating for a test
point, say x = 1, with log on
calculator & find a value that is
not shown. Observe that the
domain of f(x) in figure is x > 3.
To verify D) is shown, recall
known graph of log (x) with
vertical asymptote at x = 0.
It is now shifted to (h, k) = (3, 0)
f(x) = log (x - 3)
with a vertical asymptote at x = 3.
Problem 13
(44)
Select the matching exponential function to the graph.
Eliminate C), D), and E), choices
since they are of form ax, a >1,
which increases & is not shown.
[y = 0.65-x = (1/.65)x = (1.538)x.]
f(x) = ax = 0.65x
Evaluate A) & B) for a test point,
say x = -3, with yx on calculator.
A): 0.32-3 = 30.5 is not on graph.
B): 0.65-3 = 3.6 is on graph.
Thus, the graph is of function f(x) = ax = 0.65x.
Problem 14
(33)
Solve the problem.
log 5(x) = - 3
One computes the exponential function 5x of both
sides (which must be equal because of the given
equation) to get:
5log5(x) = 5-3 = 1/53 = 1/125.
Then, one recalls that aloga(x) = x.
Hence,
x = 1/125.
Problem 15
(33)
Solve the problem.
log 5(25) = x
Without a calculator one notes,
x = log5(25) = log5(52) = 2 log5(5) = 2.
Or, evaluate one, or other, change-of-base formula:
log5(25) = log(25)/log(5),
log5(25) = ln(25)/ln(5).
Problem 16
(72)
The formula D = 6e-0.04h can be used to find the
number of milligrams D of a certain drug in a
patient's bloodstream h hours after the drug has
been given. When the number of milligrams
reaches 2, the drug is to be given again. What is
the time between injections?
Rearrange
2 = 6e-0.04h ___________
1/e-0.04h = 6/2 or e0.04h = 3 _____
ln(e0.04h ) = 0.04h ln(e) = 0.04h = ln(3)
h = ln(3)/0.04 = 27.47 hours
Problem 17
Solve the problem.
pH = -log10[H+]
Find the pH, if the [H+] = 5.2 x 10-3.
Use your calculator.
___________ pH = -log10[5.2 x 10-3]
___________
= -(-2.284) _
__________
= 2.28
(39)
Problem 18
(56)
Solve the equation.
log 5x = log 4 + log (x - 3)
Compute the exponential function 10x of both sides
(which must be equal because of the given equation)
and use aloga(x) = x to get:
10log 5x = 10[log 4 +log(x –3)] = 10log 4 10log(x –3)]
5x = 4(x - 3)
__________ 5x - 4x = - 12________
x = -12 Note: testing => log
arguments, 5x= -60 & x-3= -15 are < 0; this is not
allowed. Thus, there is no solution. _
Problem 19
(56)
Solve the equation.
logy 8 = 5
Compute the exponential function yx of both sides
(which must be equal because of the given equation)
and use aloga(x) = x to get:
ylogy8 = 8 = y5
y5 = 8
__________
y = 81/5________
{81/5 }
Problem 20
(100)
Find the value of the expression.
Let logbA = 3 and logbB = -4.
Find logb[(AB)1/2].
logb[(AB)1/2] = (1/2)logb(AB) =
_
(1/2)[logb(A)(+ logb(B)] =
(1/2)[ (3)(+ (-4) ] = - 1/2