Math 109
Final Review
1.
Sketch the graph of f HxL = 3 - 2-x-1
The graph of this function will look like the graph of y = 2 x except reflected over the x-axis, shifted up 3,
reflected over the y-axis, and shifted left 1. (Note: The function needs to be rewritten as f HxL = 3 - 2-Hx+1L to
find the horizontal shift.) The graph has a horizontal asymptote at y = 3.
4
2
-10
-5
5
log3 81
HbL
5log5 3
-4
5. Use a change of base in order to find the decimal approximation for log7 H10L . Give the result correct to
four decimal places.
lnH10L
log7 H10L = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ º 1.1833
lnH7L
Find the domain of the following logarithmic function. gHxL = log5 H4 - x2 L
4 - x2 ¥ 0
H2 - xL H2 + xL ¥ 0
+
<---------------[------------------]-------------------->
-2
2
check -3
check 0
check 3
+++
-+
6.
-6
-8
-10
2.
Sketch the graph of the function f HxL = -3-x+2 + 1. Justify your answer.
This will look like the graph of y = 3 x , except reflected over the x-axis, shifted up 1, reflected over the y-axis,
and shifted right 2. (Note: The function needs to be rewritten as f HxL = -3-Hx-2L + 1 to find the horizontal
shift.) The graph has a horizontal asymptote at y = 1.
So the domain is @-2, 2D.
7.
4
2
Use the following graph of the logarithmic function f(x) to answer parts a and b.
2
-5
5
10
1
-2
1
3
5
7
9
11 13 15 17 19 21 23 25
-1
-4
3.
log3 81 = log3 H34 L = 4
5log5 3 = 3
10
-2
-10
HaL
Find the exponential function f HxL graphed below.
-2
18
16
14
HaL
12
10
6
4
2
The graph passes through the points H0, 2L, H1, 6L, and H2, 18L. Notice 2, 6, and 18 are all divisible by 2. If we
divide each by 2, we get 1, 3, and 9. These are all powers of 3. So we have the points
H0, 2 ÿ 30 L, H1, 2 ÿ 31 L, and H2, 2 ÿ 32 L. So f HxL = 2 ÿ 3x .
-2
4.
-1
Simplify the following.
Find b where f HxL = logb x.
Since the point H5, 1L is on the graph, we can use this to find b.
1 = logb 5
8
1
2
Thus f HxL = log5 x.
b1 = 5
b=5
HbL
Find f -1 HxL.
f -1 HxL
8.
=
11.
5x
Solve
Solve
equation.
4 H1 + 75 x L = 9.
4 H1 + 75 x L = 9
the
1 + 75 x
=
5x
=
7
6 = x2 + x
0 = x2 + x - 6
0 = Hx + 3L Hx - 2L
x = -3, 2
Checking these we see that x = -3 is extraneous, so the only solution is x = 2.
ÅÅÅÅ94
ÅÅÅÅ54
5
5 x ln7 = lnH ÅÅÅÅ
L
4
12.
lnI ÅÅÅÅ5 M
Solve
can
use
the
definition
of logarithms to
4 H1 + 75 x L = 9
do
the
2 Hx-1L
log3 H ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ L = 1
x
2 Hx-1L
31 = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ
x
3x= 2x-2
x = -2
But checking we see that x " -2, so this answer is extraneous and so there is no solution.
5
75 x = ÅÅÅÅ
4
log7 H ÅÅÅÅ54 L = 5 x
log I ÅÅÅÅ5 M
13.
x = ÅÅÅÅÅÅÅÅ75ÅÅÅÅÅ4ÅÅÅÅÅ
Note: These are the same answer. You can use the change of base formula to justify this.
Solve
the
equation
log x+2 4 = 2
Rewrite the expression in a form with no logarithms of products, quotients, or powers.
è!!!!
è!!!
x 3
lnI ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅ M = lnIx 3 M - lnHx2 + x - 2L
x2 +x-2
è!!!
= ln x + ln 3 - lnHx + 2L Hx - 1L
= ln x + ÅÅÅÅ12 ln3 - HlnHx + 2L + lnHx - 1LL
= ln x +
ÅÅÅÅ12
log2 Hx + 5L = 3
Solve for x. Use algebraic methods.
HaL
log2 Hx + 5L = 3
log x+2 4 = 2.
Hx + 2L2 = 4
x + 2 = !2
x = -2 ! 2
x = -4, 0
Checking these we see that x = -4 is extraneous and so the only solution is x = 0.
10.
equation.
log3 Hx - 1L - log3 x + log3 2 = 1
log3 Hx - 1L - log3 x + log3 2 = 1
following.
9
1 + 75 x = ÅÅÅÅ
4
9.
the
x-1
log3 H ÅÅÅÅ
ÅÅÅÅÅÅ L + log3 2 = 1
x
4
x = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ
5 ln7
you
equation
log6 x + log6 Hx + 1L = 1.
log6 x + log6 Hx + 1L = 1
log6 HxHx + 1LL = 1
61 = xHx + 1L
5
lnH75 x L = lnH ÅÅÅÅ
L
4
Or
the
ln3 - lnHx + 2L - lnHx - 1L
HbL
23 = x + 5
8= x+5
x=3
log4 H2 xL + log4 Hx - 2L = log4 Hx + 3L
log4 H2 xL + log4 Hx - 2L = log4 Hx + 3L
log4 H2 xHx - 2LL = log4 Hx + 3L
log4 H2 x2 - 4 xL = log4 Hx + 3L
è!!!!
lnI ÅÅÅÅÅÅÅÅ
ÅÅÅÅ3ÅÅÅÅÅÅ M
x2 +x-2
x
2 x2 - 4 x = x + 3
2 x2 - 5 x - 3 = 0
Hx - 3L H2 x + 1L = 0
x = 3, - ÅÅÅÅ12
1
Checking these we'll find that x = - ÅÅÅÅ
is
extraneous
and
so x = 3 is the solution.
2
HcL
‰ x x2 + 2 x ‰ x + ‰ x = 0
e x x2 + 2 x e x + e x = 0
ex Hx2 + 2 x + 1L = 0
ex Hx + 1L2 = 0
x = -1
14. The half-life of Strontium-90 is 28 years. How long will it take a 50mg sample to decay to a mass of
32mg?
We'll use the model A = P ek t , to answer this question.
Assuming that we start with 100% of the mass or P = 1 and end with 50% of the mass or A = .5, we get the
following
equation.
.5 = 1 ekH28L
lnH.5L = 28 k
lnH.5L
k = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ º -.0248
28
Now start with P = 50 mg and use the value for k above with A = 32 mg and get the following equation that
can
be
solved
for
t.
32 = 50 e-.0248 t
.64 = e-.0248 t
lnH.64L = -.0248 t
lnH.64L
t = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ º 18.03 years
-.0248
14. The half-life of Strontium-90 is 28 years. How long will it take a 50mg sample to decay to a mass of
32mg?
We'll use the model A = P ek t , to answer this question.
Assuming that we start with 100% of the mass or P = 1 and end with 50% of the mass or A = .5, we get the
following
equation.
.5 = 1 ekH28L
lnH.5L = 28 k
17. Solve the equation graphically. x3 = ex+1 - 3
First plot y = x3 and y = ex+1 - 3 on your calculator.
2
1
-3
-2
-1
lnH.5L
k = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ º -.0248
28
Now start with P = 50 mg and use the value for k above with A = 32 mg and get the following equation that
can
be
solved
for
t.
32 = 50 e-.0248 t
.64 = e-.0248 t
lnH.64L = -.0248 t
15. $3000 is invested at an annual percentage rate of 8.5%. How long will it take the account to reach $5000
if the interest is compounded monthly? Use algebraic methods to solve.
We'll
use
the
formula
A = PH1 +
P = 3000, r = .085, A = 5000, and n = 12.
.085
5000 = 3000 H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
with
-2
-3
-4
lnH.64L
t = ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅ º 18.03 years
-.0248
ÅÅÅÅnr Ln t ,
Then use the intersection feature to find the indicated points. This gives the answers x º -1.31, .099.
18.
The initial count in a bacteria culture was 241. After 5 hours, the count was 1000.
HaL
Assume the growth is exponential. Find a formula for the number of bacteria nHtL after t hours.
We'll use the model nHtL = n0 ek t .
12 t
5
.085
ÅÅÅÅ
= H1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
3
12
.085
lnH ÅÅÅÅ53 L = lnIH1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
12 t
M
.085
lnH ÅÅÅÅ53 L = 12 t lnH1 + ÅÅÅÅ
ÅÅÅÅÅÅ L
12
lnI ÅÅÅÅ5 M
12 lnI1+ ÅÅÅÅ12
ÅÅÅÅÅÅÅÅ M
3
t = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅ º 6.03 years
.085
16. $3000 is invested into an account paying 4% annual interest compounded bimonthly. How long will it
take for the account to reach $3500?
nt
We'll use the formula A = PH1 + ÅÅÅÅnr L , with A = 3500, P = 3000, n = 6, r = .04, and then solve for t.
.04
3500 = 3000 H1 + ÅÅÅÅ
ÅÅÅÅ L
6
6t
7
lnH ÅÅÅÅ
L
6
7
lnH ÅÅÅÅ
L
6
= H1 +
.04 6 t
ÅÅÅÅ
ÅÅÅÅ L
6
.04
= lnIH1 + ÅÅÅÅ
ÅÅÅÅ L M
6
6t
.04
= 6 tHlnH1 + ÅÅÅÅ
ÅÅÅÅ LL
6
lnI ÅÅÅÅ7 M
6 lnI1+ ÅÅÅÅ6ÅÅÅÅÅ M
6
t = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅ
.04ÅÅÅÅÅ º 3.87 years
17. Solve the equation graphically. x3 = ex+1 - 3
First plot y = x3 and y = ex+1 - 3 on your calculator.
2
1
-3
-2
-1
1
-1
-2
-3
-4
Then use the intersection feature to find the indicated points. This gives the answers x º -1.31, .099.
Then use n = 1000, n0 = 241, and t = 5 to find k.
1000 = 241 ekH5L
1000
ÅÅÅÅÅÅÅÅ
ÅÅÅÅ
241
1000
lnH ÅÅÅÅÅÅÅÅ
Å
ÅÅ
ÅL
241
1000
lnH ÅÅÅÅÅÅÅÅ
Å
ÅÅ
ÅL
241
12 t
ÅÅÅÅ76
1
-1
So nHtL = 241 e
HbL
.285 t
= e5 k
= lnHe5 k L
=5k
1000
lnI ÅÅÅÅÅÅÅÅ
ÅÅÅÅÅ M
k = ÅÅÅÅÅÅÅÅÅ5241
ÅÅÅÅÅÅÅÅÅÅ º .285
.
Predict the number of bacteria after 8 hours.
nH8L = 241 e.285 H8L º 2349