Download Handout 6 - UGA Math Department

Name:
MATH 1113 Precalculus
Eric Perkerson
Date: October 14, 2014
Handout No. 6
Problem 2 Evaluate
s
loga
ax12
y6
!
given that loga (x) = −8.2 and loga (7) = 4.5.
Expand using properties of logarithms:
s
!
ax12
loga
= loga
y6
ax12
y6
1/2 !
12 ax
1
= loga
2
y6
1
loga (a) + loga (x12 ) + loga (y −6 )
=
2
1
= (1 + 12 loga (x) − 6 loga (y))
2
and substitute in the values of loga (x)and loga (y) to solve.
Problem 3 v.1 Solve the equation:
log13 (2x − 3) = log13 (12) − log13 (3)
Solution:
log13 (2x − 3) = log13 (12) − log13 (3)
log13 (2x − 3) = log13 (12/3)
log13 (2x − 3) = log13 (4)
2x − 3 = 4
2x = 7
x = 7/2
Problem 3 v.2 Solve the equation for x in terms of the positive real variable a:
loga (x − 3) = 2
Solution:
loga (x − 3) = 2
x − 3 = a2
x = a2 + 3
Problem 3 v.3 Solve the equation:
loga (x) = loga (3 − 3x)
Solution:
loga (x) = loga (3 − 3x)
x = 3 − 3x
4x = 3
x = 3/4
Problem 3 v.4 Solve the equation:
log3 (7x + 1) = 4
Solution:
log3 (7x + 1) = 4
7x + 1 = 34
7x = 81 − 1
x = 80/7
Problem 4 v.1 Solve the equation for x:
a− loga (x) = .01
Solution:
a− loga (x) = .01
a− loga (x) = 1/100
aloga (x
−1
x
)
= 1/100
−1
= 1/100
1
= 1/100
x
x = 100
Problem 4 v.2 Find all solutions of the equation
ln(x) + ln(x + 4) = ln(33/4)
Solution:
ln(x) + ln(x + 4) = ln(33/4)
ln(x(x + 4)) = ln(33/4)
x2 + 4x = 33/4
x2 + 4x − 33/4 = 0
x=
−4 ±
√
16 + 33
2
−4 ± 7
2
3 −11
x= ,
2 2
x=
But remember that x = −11/2 is not in the domain of ln(x), thus the only solution to the equation is
x = 3/2.
Problem 4 v.3 Find all solutions of the equation
log6 (x − 5) + log6 (x) = 2
Solution:
log6 (x − 5) + log6 (x) = 2
log6 ((x − 5)x) = 2
x2 − 5x = 62
x2 − 5x − 36 = 0
5±
p
25 + 4(36)
p 2
5 ± 25 + 4(36)
x=
2
5 ± 13
x=
2
18 −8
,
x=
2 2
x = 9, −4
x=
But remember that x = −4 is not in the domain of log6 (x), thus the only solution to the equation is
x = 9.
Problem 4 v.4 Find all solutions of the equation
log2 (x + 3) = log2 (x − 3) + log5 (25) + 7log7 (3)
Solution:
log2 (x + 3) = log2 (x − 3) + log5 (25) + 7log7 (3)
log2 (x + 3) − log2 (x − 3) = 2 + 3
x+3
log2
=5
x−3
x+3
= 25
x−3
x + 3 = 32(x − 3)
x + 3 = 32x − 96
−31x = −99
x = 99/31
Problem 5 Solve the equation
23x+16 = 56−13x
Solution:
log2
23x+16 = 56−13x
23x+16 = log2 56−13x
3x + 16 = (6 − 13x) log2 (5)
3x + 16 = 6 log2 (5) − 13 log2 (5)x
3x + 13 log2 (5)x = 6 log2 (5) − 16
x(3 + 13 log2 (5)) = 6 log2 (5) − 16
x=
Problem 6 v.1
6 log2 (5) − 16
3 + 13 log2 (5)
Find all solutions of the equation:
e2x 10ex + 25 = 0
This is a quadratic in ex problem:
e2x 10ex + 25 = 0
(ex )2 − 10(ex ) + 25 = 0
x
e =
10 ±
p
100 − 4(25)
2
10
2
ex = 5
ex =
x = ln(5)
Problem 6 v.2
Find the exact solution, using algebra and logarithms:
6x + 216(6−x ) = 42
This is a hidden quadratic in 6x problem, so start by multiplying both sides of the equation by 6x :
6x + 216(6−x ) = 42
(6x )2 + 216 = 42(6x )
(6x )2 − 42(6x ) + 216 = 0
x
6x
6x
6x
6x
p
422 − 4(216)
√ 2
42 ± 900
=
2
42 ± 30
=
2
72 12
=
,
2 2
= 36, 6
6 =
42 ±
x = log6 (36), log6 (6)
x = 2, 1
Problem 6 v.3
Find all solutions of the equation:
ex − 56e−x = −1
This is a hidden quadratic in ex problem, so start by multiplying both sides of the equation by ex :
ex − 56e−x = −1
(ex )2 − 56 = −ex
(ex )2 + ex − 56 = 0
−1 ±
p
1 + 4(56)
√2
−1
±
225
ex =
2
−1 ± 15
x
e =
2
ex = 14/2, −16/2
x
e =
x = ln(7), ln(−8)
But we remember that −8 is not in the domain of ln, so the only solution is ln(7).
Problem 7 v.1 Use natural logarithms to solve for x in terms of y:
y=
We can rearrange to get
ex − e−x
4
4y = ex − e−x
Now this is a hidden quadratic in ex problem, so we multiply both sides of the equation by ex :
4y = ex − e−x
4yex = (ex )2 − 1
0 = (ex )2 − 4yex − 1
p
4y ± (4y)2 + 4
x
e =
2
p
Be careful here: Because (4y)2 + 4 > (4y)2 , we have that (4y)2 + 4 > 4y, so
p
4y − (4y)2 + 4
<0
2
thus the only solution is
x = ln
4y +
p
(4y)2 + 4
2
!
Problem 7 v.2 Use natural logarithms to solve for x in terms of y:
y=
e4x + e−4x
e4x − e−4x
Solution:
e4x + e−4x
e4x − e−4x
4x
−4x
y(e − e
) = e4x + e−4x
y=
ye4x − ye−4x ) = e4x + e−4x
ye4x − e4x = e−4x + ye−4x )
e4x (y − 1) = e−4x (1 + y)
e4x /e−4x = (1 + y)/(y − 1)
e8x = (1 + y)/(y − 1)
1+y
8x = ln
y−1
1+y
x = (1/8) ln
y−1