Lesson 6
Exponential and Logarithmic Fu tions Lesson 6
Exponential and Logarithmic Functions Exponential functions are of the form y = ax where a is a constant greater than zero and not equal to one and x is a variable. Both y = 2x and y = ex are exponential functions. The function, ex, is extensively used in calculus. You should memorize its approximate value when x = 1. (e1 ≈ 2.718) You should also be able to quickly graph y = ex without the aid of a calculator. A simple graph of y = ex is shown below.
Figure 1
3 2 1 –1
1 2 3
y = ex
Logarithmic FunctionS
The equation y = log ax is the same as ay = x. The inverse of the exponential function is y = ax. In this course we will restrict our study of logarithms to log base e which will be written as ln(x). The equation y = ln(x) is the inverse function of y = ex. Notice that the graph of ln(x) is a reflection of graph of ex around the line y = x. You should be able to quickly sketch from memory y = ln(x). It will also be important to remember the basic logarithm rules listed on the next page. EXPONENTIAL AND LOGARITHMIC FUNCTIONS - LESSON 6
59
Figure 2
y = ln(x)
I.
ln(1) = 0
II.
ln(e) =1
III. ln(ex) = x
IV. eln(x) = x
V.
VI. Quotient: ln(x/y) = ln(x) – ln(y)
VII. Power: ln(xa) = a ln(x)
Product: ln(xy) = ln(x) + ln(y)
Remember that the natural log of a negative number is undefined. Some books specify ln(x) as ln |x|. We will use ln(x) for this book. Be careful to use only positive, non-zero values for x when employing the natural log function.
60
LESSON 6 - EXPONENTIAL AND LOGARITHMIC FUNCTIONS
CALCULUS
The natural log function can be used to free variable exponents from their exponential functions. Conversely, the exponential function can do the same for the natural log functions. Example 1 Solve for x.
e2x = 1
Taking ln of both sides:
( )
ln e2x = ln (1) 2x = 0 x=0
checking e2(0) = e0 = 1
Example 2 Solve for x.
ln(x + 5) = 0
Use each side of the equation as the exponent for e. eln(x + 5) = e0 x + 5 = 1; so x = –4
Sometimes the equations are complex and we need to use substitution to solve them. See example 3 on the next page.
CALCULUS
EXPONENTIAL AND LOGARITHMIC FUNCTIONS - LESSON 6
61
Example 3 Solve for x.
e2x – 4ex + 3 = 0
Substituting u = ex, u2 – 4u + 3 = 0. Factoring, we get (u – 3)(u – 1) = 0. Replacing u with ex, we get (ex – 3)(ex –1) = 0. Solving each factor, we get: ex = 3; ex = 1. Taking the ln of both sides:
( )
ln ex = ln (3)
x = ln (3)
( )
ln ex = ln (1) x=0
Example 4 Draw the graph of y = 2ex and its inverse. y = 2ex x = 2e
y
y = 2ex switch variables
1 x = ey 2
( ) ( )
y=x
( )
ln 1 x = ln ey 2 ln 1 x = y 2
62
( )
( )
f −1( x ) = ln 1 x 2
f −1( x) = ln 1 x 2
LESSON 6 - EXPONENTIAL AND LOGARITHMIC FUNCTIONS
CALCULUS
lesson practice
6A
Answer the question. 1.
x Draw the graph of y = e . Find the inverse function. Graph it. 3
2. Draw the graph of y = 2ex. Find the inverse function. Graph it.
3. Solve for x.
A.
e2x + 1 = 1
B.
2e3x = e0
C.
0 = ln(2x +5)
calculus Lesson Practice 6A
45
LESSON PRACTICe 6A
D.
ln(x) + ln(5) = 6
4. Solve for x. (Hint: Substitute and factor.)
46
A.
e2x – 5ex = –6
B.
2e2x + 7ex = 4
calculus
6B
lesson practice
Answer the question. 1.
2.
Draw the graph of y = ex + 1. Find the inverse function. Graph it.
Draw the graph of y = e
x 2
. Find the inverse function. Graph it.
3. Solve for x.
A.
ex + ln(3) = 2
B.
ex + 1 = e2x – 2
calculus Lesson Practice 6B
47
LESSON PRACTICe 6B
C.
ln(x2 + 3x + 5) = ln(1 – x)
D.
ln x 2
( )=3
4. Solve for x.
48
A.
2ln2(x) + 3 = 7ln(x)
B.
e2x = 2ex
calculus
lesson practice
6C
Answer the question. 1. Draw the graph of y = 2x2. Find the inverse function. Graph it.
Solve for x. 2. e4x = e
3. ln(3x – 1) = 1
calculus Lesson Practice 6C
49
LESSON PRACTICe 6C
4. e2x – 7ex + 10 = 0
5. ln2(x) = 2 ln(x)
6. e2x – 3ex + 2 = 0
50
calculus
lesson practice
6D
Solve for x. 1. e2x + 2 = 5
2. 2e2x + 5ex = 3
3. ln(x + 2) = 2
calculus Lesson Practice 6D
51
LESSON PRACTICe 6D
4. ln(x + 1) + ln(4) = 3
5. Solve for x: ln(2x – 4) = 2.
6. Draw the graph of y = e3x. Find the inverse function. Graph it.
52
calculus
6
test
Circle your answer. 1. Simplify ln (9) 2
A.
B.
=
ln(4.5) 1 ln(4.5) 2
C.
ln(3)
cannot be simplified
D.
2. Solve for x: ln(x) – ln(4) = 2. 1 2 e 4
A.
B.
e4
C.
e2
4e2
D.
( 3 ) is the same as:
3. ln 6
A.
2
B.
ln(18)
C.
ln(3)
ln(6) – ln(3)
D.
4. Find the inverse function: f(x) = ln(x – 2).
A.
f –1(x) = ln(x + 2)
B.
f –1(x) = ex + 2
C.
f –1(x) = 2ex – 2
f –1(x) = 2ex
D.
5. Simplify ln ( 2 ) + ln ( 10 ) . 12
)
C.
ln 2 5
)
cannot be simplified
A.
B. D.
calculus Test 6
ln
(
20
(
17
Test 6
6. Solve for x: ln2(x) – 5 ln(x) = –4
A.
x = e, e4
B.
x = ln(4), ln(5)
C.
x = e 5, e
x = ln(4), e
D.
7. e X and ln(x) are inverse functions. The graph of y = ln(x) is the reflection of the graph of y = e x around the: A. B. C. D.
x-axis y-axis origin line y = x
8. Solve for x: e2x = 3ex
A.
x = ex – 3
B.
x = ln(3)
C.
x = ln(3) and x = 0
x=0
D.
9. Solve for x: ln(2) + ln(x) = 7 e2
A.
x= 7
B.
x = 7e2
C.
x = 2e7
7 x= e
D.
2
10. Solve for x: e3x – 1 = 1
18
A.
B.
1 3 –
1 3
C.
3
e3
D.
calculus
3x = ey
phase shift = −π period = 2π = π 2 5. 6. 7.
( )
ln (3x) = ln ey
( ) tan( 3π ) = tan (135º ) = –1 4
csc 2π = csc(120º ) = 3
= 2 3 3 3
cos (2x) = 0 for x in [0, π] cos(θ) = 0 when θ = π , 3π , 5π etc 2 2 2 3 π π 5 2x = 2x = 2x = π 2 2 2 3 5 π π x= x= π x= 4 4 4 x = π and 3π 4 4
8.
( ) [
]
tan 1 x = 0 0, π 2 2 tan ( θ ) = 0 when θ = 0, π, 2π, 3π etc. 1x=0 2 x=0
ln (3x) = y
2
f −1( x) = ln (3x)
L e ss o n Pra c ti c e 5 D - L E SS ON PRA CTI CE 6A
Lesson Practice 6A
y ) ey (reverse variables x = ex =(2Switch variables s) 3 y ln ( x) = ln 2e 3x = (ey) ln x = ln2 + y ln (3x) = ln ey ln ( x) − ln (2) = y ln (3x) −=1y f ( x) = ln ( x) − ln (2) f −1( x) = ln (3x) note: This problem is the same as example 4 in the e instruction manual but solved differently. = 2ex 2. Bo othysolutions are correct.
(
[
]
Lesson Practice Lesson Practice6A 6A
y=x ln ( x) − ln (2) = y 2x+1 3.. A. e =1 f −1( x) = ln ( x) − ln (2) 2x+1 = ln (1) ln e note: This problem is the same as 2x + 1 = 0 y = 2ex 4 in the example e instruction 2x = −1 manual but solved differently. x = − 1 –1 Bo oth solutions2fare correct. (x)= ln(x) – ln(2)
(
B.
( Switch variabless)
3x = ey
3..
ln (3x) = ln (ey )
ln (3x) = y f −1( x) = ln (3x)
2.
x = 2e
y=x y
B.
(reverse variables)
D.
) = ln(3x) ln ( x) = lnf(2−1 ey(x) ln ( x) = ln2 + y ln ( x) − ln (2) = y f −1( x) = ln ( x) − ln (2)
note: This problem is the same as example 4 in the e instruction manual but solved differently. Bo oth solutions are correct.
( )
calculus
3..
ln ( x ) − ln (2) = ln x 2
A.
(
e2x+1 = 1
)
A. c.
ex fy(x) = 2=ex3
ln e2x+1 = ln (1)
() )
x = 2ey (reverse variables) ln ( x) − ln (2) = ln x ln ( x) = ln 2ey 2 ln ( x) = ln2 + y
x y= e 3
y x= e 3
)
( )
(
1x=π 2 x = 2π
x = 0 is the only answer in 0, π 2
1.
x y= e x y =32e
1. 2.
c.
4.
A.
D.
)
( )
0 x e3(x2)==eln ln ( x ) −2ln 2 ln 2e3x = ln e0 ln (2) + 3x = 0 3x = − ln (2) e2x+1 = 1 ( ) x = − ln 2 ln e2x+1 = ln (1) 3 =+ 0 5) =xln+(1 02 2x 2x( = −1 ) e0 = eln 2x+5 1 1 = 2xx+=5− 2 −4 = 2x 2−e43x = e0 x = 3x = −2 0 ln 2e2 = ln e ln ln (2 x) ++ 3 lnx(5=) 0= 6 ln3(x 5x=) −=ln 6(2) ( ) ( ) elnx5x= −= ln e6 2 3 5x = e6 0 = ln (2x + 5) 6 e or 1 e6 ( x )= e0 = eln 2x+5 5 5 1 = 22xx + 5 x e − 5e = −6 Let y = ex. −4 = 2x 2 y − 5y + 6 = 0 x = −4 = −2 ( y − 2) 2( y − 3) = 0
(
(
)
( )
)
( )
)
(
269
s o luti o n s lny( x −)2+=ln0(5) = 6 y −3= 0
2x) = 6 yln=(5 (5x) ln 6 exe= 2 = e
( )
6
y=3 ex = 3
( )
5(x2)= e ln ex = ln ln ex = ln (3) 6 x = ln (x2)= e or 1 e6 x = ln (3) 5 5
B.
2e3x = e0
(
)
( )
ln 2e3x = ln e0 ln (2) + 3x = 0 Le ss o n Practice 6 A - LE3 SS (2) TI CE 6B = − PlnRAC xON ( ) x = − ln 2 3 c.
Lesson 6B Lesson Practice 6B
0 = ln (2x + 5)
(
D.
x = ey +1 (reverse variables)
(
ln ( x) − 1 = y f −1( x) = ln ( x) − 1
ln (5x) = 6
Lesson 6B
( )
eln 5x = e6
A.
e
− 5e = −6
2. x
Let y = e .
y
( y − 2) ( y − 3) = 0 y −2= 0
y −3= 0
y=2
y=3
( ) = ln(2) x
ln e
x = ln (2)
B.
2e2x + 7ex = 4
ex = 3
( ) = ln(3) x
ln e
f
2.
Let y = ex.
y
3.
A.
y = −4
y= 1 2 x e = 1 2
x
()
f
e = −4
ln (ex ) = ln ( –4) error: no solu ution
B. 3.
A.
c. B.
D.
270
so luti ons
c.
(reverse variables)
(( )) ) y
ex+ln 3 = 2 ln ( x) = ln ( e) 2 ln ex+ln 3 = ln (2) y ) =+ ln ln ( xx ( ) ( ) 2 3 = ln 2
(
x = ln (2) − ln (3)
2 ln ( x) = y
y+4=0
2y = 1
( )
y=e2
x=e2
(2y − 1) ( y + 4) = 0
ln ex = ln 1 2 x = ln (1) − ln (2) x = 0 − ln (2) x = − ln (2)
(x) = 2 xln (x)
–1
x = ln (3)
2y2 + 7y − 4 = 0 2y − 1 = 0
)
( )
y − 5y + 6 = 0
e =2
(
ln ( x)( −)1 = y 2 ln x = ln e f −1( x) = ln ( x) − 1 y ln ( x) = 2 2 ln ( x) = y
2
x
y=x ) x = ey +1 (reverse variables y=e2 ln ( x) = 2.7 lny ey +1 f −1( x ) = ln(x) – 1 y = ex( +x1)= e 2 (reverse variables) ln x = y + 1 x
6 x = e or 1 e6 5 5
4.
y = ex+1
1.
5x = e6
x
)
ln ( x) = ln ey +1 ln ( x) = y + 1
ln ( x) + ln (5) = 6
2x
y = ex+1
1.
)
e0 = eln 2x+5 1 = 2x + 5 −4 = 2x x = −4 = −2 2
()
lnx( x=) ln 2 3
( x) = 2
–1
(
ex+1 = e2x−2
)
(
)
ln exx++1ln=(3)ln e2x−2 y=x =2 ex x + 1 = 2x − 2 ( ) y ln = −eexx2++ln13= −2 =2 ln (2) f –1 (x) = 2 ln(x) ( ) x +−ln x =3 −=3 ln (2) x =x3= ln (2) − ln (3)
(
)
)(
(
)
2 (1− x) ln x2 + 3xx+=5ln= ln 3 x2 + 3x + 5 = 1− x x+1 2x−2 xe2 + 4=xe+ 4 = 0
(
)
(
)
2x−2 ln( xe+x+21) (=xln + 2e) = 0 x + 1 =x2+x 2− =2 0 −x + 1 = −2 2x = −2 −x = −3 ln x x==33 2 lnlnx2x + 3x + 5 = ln (1− x) Calculus e 22 = e3 x + 3x + 5 = 1− x x 2 = e3 x2 + 4x + 4 = 0 (x + 2) (x +32) = 0 x = 2e x+2= 0 x = −2
( ) (( )
)
f
( x) = 2
–1
ln ( x) = 3 ( ) eln x = e3
y= 1 2 ln ( x) = 1 2
y 2 2 ln ( x) = y ln ( x) =
ln ( x)
x = e3
1 L e ss o n Pra ( ) c ti c e 6 B - L E SS ON PRA CTI CE 6C
eln x = e 2
3.
A.
x+ln(3)
(
e
( ))
B.
=2
x=
e
2x
= 2ex
e
y2 = 2y
ln ex+ln 3 = ln (2) x + ln (3) = ln (2)
y2 − 2y = 0
x = ln (2) − ln (3)
( y ) ( y − 2) = 0
()
x = ln 2 3 B.
x
( ) = ln(0)
ln e
ln (ex+1) = ln (e2x−2)
undefined
)
ln x2 + 3x + 5 = ln (1− x)
s) x = 2y2 (reverse variables x = y2 2
( )
f −1( x) = ±
( ) = e3
2.
x = e3 2 x = 2e
2 ln2 ( x) + 3 = 7 ln ( x)
Let y = ln ( x)
3.
(2y – 1)( y – 3) = 0 y −3= 0
( ) eln x = x=
e
e2x = 2ex
)
4.
y=3 ln ( x) = 3 ( ) eln x = e3
e2x − 7ex + 10 = 0
y −2= 0
y =5
y=2
ex = 5
ex = 2
( )
x = ln (5)
5.
( )
ln ex = ln (2) x = ln (2)
271 n sy = let ln ( x) ln2 ( x) = 2 ln ( x) s o luti o y2 = 2y y2 − 2y = 0
y=0
y −2= 0
ex = 0
y=2
( y ) ( y − 2) = 0
ln ex = ln (0)
ex = 2
y=0
( )
let y = ex
y −5= 0
ln ex = ln (5)
Lett y = ex
( y ) ( y − 2) = 0
x 2
( y − 5) ( y − 2) = 0
x=e
y2 − 2y = 0
)
y2 − 7y + 10 = 0
3
y2 = 2y
y=x
ln (3x − 1) = 1
(
2y2 − 7y + 3 = 0
1 e2
e4x = e
eln 3x−1 = e1 3x − 1 = e 3x = e + 1 –1 x = e + 1f (x) = ± 3
2y2 + 3 = 7y
2Y = 1 y= 1 2 ln ( x) = 1 2
(
x 2
ln e4x = ln (e) 4x = 1 x= 1 y = 2x2 4
3
calculus
x =y 2
±
ln x e 2
B.
y = 2x2
1.
ln x = 3 2
2y − 1 = 0
ln ex = ln (2) x = ln (2)
Lesson Practice Lesson Practice6C 6C
x + 3x + 5 = 1− x
A.
( )
x = ln (0)
x2 + 4x + 4 = 0 (x + 2) (x + 2) = 0 x+2= 0 x = −2
4.
ex = 2
x
2
D.
y=2
e =0
ex+1 = e2x−2
(
y −2= 0
y=0
x + 1 = 2x − 2 −x + 1 = −2 2 −x = −3 x=3 c.
Lett y = ex
y−2= 0
s x = 2y reverse variables x = y2 2 x =y 2
±
Le ss o n Practice 6 C - LE SSON P RAC TI CE 6D
f −1( x) = ± 2.
(
x 2
Lesson Practice Lesson Practice6D 6D
e4x = e
)
ln e4x = ln (e) 4x = 1 x= 1 4 3.
1.
2x + 2 = ln (5)
2x = ln (5) − 2 (5) x = ln − 2 2
ln (3x − 1) = 1
(
)
2e2x + 5ex = 3
2.
2y2 + 5y − 3 = 0
e2x − 7ex + 10 = 0
(2y − 1) ( y + 3) = 0
let y = ex
y2 − 7y + 10 = 0
2y − 1 = 0
( y − 5) ( y − 2) = 0
2y = 1
y −5= 0
y −2= 0
y =5
y=2 ln ex = ln (2) x = ln (2)
x = ln (5)
ln2 ( x) = 2 ln ( x)
let y = ln ( x)
y = 2y y − 2y = 0
3.
( y ) ( y − 2) = 0
0
x=e x =1 6.
x + 2 = e2 x = e2 − 2
y=2 ln ( x) = 2 ( ) eln x = e2
4.
ln ( x + 1) + ln ( 4) = 3 ln ( 4 ( x + 1)) = 3
ln ( 4x + 4) = 3 ( ) eln 4x+ 4 = e3
x = e2
e2x − 3ex + 2 = 0
4x + 4 = e3
(ex − 1) (ex − 2) = 0 ex − 1 = 0 ex = 1 x = ln (1) x = 0,ln (2)
undefined
ln ( x + 2) = 2 ( ) eln x+2 = e2
y−2= 0
ln ( x) = 0 ( ) eln x = e0
ex = −3 x = ln ( −3)
x = ln 1 2 x = ln1− ln (2) x = 0 − ln (2) x = − ln (2)
2
2
y = −3
()
( )
( )
ln ex = ln (5)
y +3= 0
y= 1 2 x e = 1 2
ex = 2
e =5
y=0
let y = ex
2y2 + 5y = 3
x
5.
)
ln e2x+2 = ln (5)
eln 3x−1 = e1 3x − 1 = e 3x = e + 1 x = e +1 3 4.
(
e2x+2 = 5
4x = e3 − 4
ex − 2 = 0 ex = 2 x = ln (2)
3 x= e −4 4
5.
ln (2x − 4) = 2 2x − 4 = e2 2x = e2 + 4 2 x= e +4 2
6.
272
so luti ons
x
y
0
1
1
e3
–1 1 3
e–3 e
Finding the inverse
Calculus
x= x = e2 − 2 5. 4.
4
x +− 14))+=ln2( 4) = 3 lnln(2( x 2x − 4 ln=( 4e(2x + 1)) = 3
6.
5.
6.
2x = el2n+( 44x + 4) = 3 2 eln(4x+ 4) = e3 x= e +4 2 4x + 4 = e3 x y 4x = e3 − 4 0
1
1
3
3
x= e −4 4
e
L e ss o n Pra c ti c e 6 D - L E SS ON PRA CTI CE 7A
Lesson Practice5A 7A Lesson Practice
(
)
lim x2 − x + 3 =
1.
x →3
lim x2 − lim x + lim 3 =
ln–(12x −e–43) = 2 2x1− 4 =ee2 3 2x = e2 + 4
x →3
x →3
x →3
9 −3+3 = 9
2 (1, 20) 4 inverse Finding x = e +the 2 3x x yy= e 3y 0 x1= e ) 3= 3y 1ln ( xe
( ) –3 –1 e y = ln x 3 1 –1 e 1( ) ln − 1 f3 ( x) = y =x e 3x 3 Finding the inverse y = e3x
)
lim 7 − 2x − x2 = 7 − 4 − 4 = −1
3.
2 lim t − 3t + 3 = 0 − 0 + 3 = 3 t →0 0+2 2 2t2 + 2
4.
x→2
z + 1 = −2 + 1 = −1 4−2 2 z2 − 2
lim
z→−2
lim
5.
x = e3y ln ( x) = 3y
θ→ π 4
tan (3θ ) = sin ( θ ) cos ( θ )
( ) = ( ) ( )
tan 3π 4 π cos π sin 4 4 −1
lim
θ→ π 4
( ) y = ln x 3 ( x) ln −1( ) f x = 3
y = ln( x) 3
(
2.
2 ⋅ 2 6.
2 2
= −1 = −2 2 4
x2 − 9 x2 − 3x
lim x →3
When x = 3, the function is undefined. Factoring we get,
1
lim
(x + 3) ( x − 3 )
(
x →3
x x−3 lim x →3
7.
x→−3
calculus
( x + 3)
lim
x→−1
lim
x→−1
x
(x + 2) ( x + 3 )
x→−3
8.
(x + 3)
= = 6 =2 3
x2 + 5x + 6 = x+3
lim
lim
)
= −1
x2 + 5x + 6 = (x + 1)
( x + 2) (x + 3) (x + 1)
s o luti o n s
=
limit does not exist 9.
lim (sin ( θ ) + 2 cos ( θ )) = θ→ 0
273
6. c : A is the sine function. B is a non-vertical line, so it is a fu unction. c is a parabola with the "c" shape so itt fails the vertical line test. D is the naturall log function. 7. B : definition of inverse 8. c : f ( x ) = 2x2
T e st 4 - TEST 6
4.
A
5.
A : tan 3π = tan135º 4 quadrant II, tangent is negative
x = 2y2 (switch variables)
definition of frequency
( )
( )
tan 3π = −1 4
x = y2 2 y=±
x 2
6. c
Definition of amplitude
f −1( x) = ±
x 2
7. B
–π
8. B
2 is the vertical shift
9. c
csc 7π = csc 315º = – 2 4
10. D
sin (2x) = 0 when x = 0, π , π, 2 3π , 2π, 5π , ... 19π , 10π 2 2 2 There are 21 solutions.
This is a graph of a parabola with the e "c" shape. It is not a function, h is in quadrants but its graph I and IV. A.
θ=π
θ = 0, π , π 2
y = 2x2
9.
θ= π 2
θ=0
c = πd c =d π d ( c) = c π
( )
Test Test 5 1.
B : The period of y = cos ( x) is 2π, so the period of y = cos (2x) will be half as large.
( )
c
2 = 1
( ) 1
1.
c
2. D
hyp 2. c : sec π = = 4 adj 3.
Test 6 Test
ln (9) = 1 ln (9) = ln 9 2 = ln 2 2
( )
2
ln x = 2 4
θ=0
x = e2 4
θ = 0, π , π 2 4.
A
x = 4e2
2θ = 2π θ=π
( ) = e2
ln x 4
e
3π, 4π etc 2θ = π θ= π 2
)
9 = ln (3)
ln ( x) − ln ( 4) = 2
sin ( y ) = 0 When y = 0, π, 2π π,
2θ = 0
(
3.
D
( )
ln 6 = ln (6) – ln (3) 3 y = ln ( x − 2)
4. B
x = ln ( y − 2) (switch va ariables) ex = eln( y −2)
definition of frequency
ex = y − 2 5.
calculus
( )
ex + 2 = y
A : tan 3π = tan135º 4 quadrant II, tangent is negative
( )
tan 3π = −1 4
6. c
Definition of amplitude
7. B
–π
f −1( x) = eX + 2
(
)
2 + ln
(
)
( = ln (
5.
c
ln
10 = ln
6.
A:
ln2 ( x) − 5ln ( x) = −4
) ( 10 ) 427 20 ) = ln (2 5 ) 2
s o luti o n s
y2 − 5y = −4 y2 − 5y + 4 = 0
( y − 1) ( y − 4) = 0
Test 5 1. D x = ln ( y − 2) (switch va ariables) Test 5 y = ln ( x − 2)
4. B
ex = eln( y −2)
1.
D
2. ) ( 10 ) 20 ) = ln (2 5 ) 2.
c
x
T e st 6 - TEST 7
e = y−2 x
e +2= y f −1( x) = eX + 2
5.
6.
c
A:
ln
(
)
2 + ln
(
)
( = ln (
10 = ln
2
Limit from the left = 1 and limit from the right = 2. They are not Limit from the left = 1 and limit equal, so the limitt does from the right = 2. They are not not exist. equal, so the limitt does not exist.
c
ln2 ( x) − 5ln ( x) = −4 y2 − 5y = −4 y2 − 5y + 4 = 0
( y − 1) ( y − 4) = 0 y − 1= 0
y−4=0
y =1
y=4
ln ( x) = 1 ( ) eln x = e1 x=e 7. 8.
D B
ln ( x) = 4 ( ) eln x = e4
c
3.
c
4. B
lim
from lesson 4
5.
D
lim
e2x = 3ex
5.
D
lim
6. B
lim
6. B
lim
7.
c
lim
7.
c
lim
x = e4
definition of an inverse
ln (e2x ) = ln (3ex )
( )
2x = ln (3) + x x = ln (3) ln (2) + ln ( x) = 7 ln (2x) = 7
1.
D
Limit from the left = 1 and limit from the right = 2. They are not equal, so the limitt does not exist.
428 2. so c luti ons
x→1
θ→ 0 θ→ 0
x→2 x→2
x→2
lim
7 x= e 2
3x − 1 = 0 3x = 1 x= 1 3
x→1
x→2
2x = e
ln (e3x−1) = ln (1)
x→2
lim
7
e3x−1 = 1
x→2
lim
( ) eln 2x = e7
Test 57 Test
lim sin ( x) = 0
x→ π
lim
2x = ln (3) + ln e
10. A
x→ π
4. B
x
9. D
3.
limit from the left and right = 0 limit from the left and right = 0 lim sin ( x ) = 0
x→2
4 =2 2 4 =2 2 = −1 = −1
lim x + 3 = 5 x→2
8..
D
8..
D
9.
A
9.
A
10. A 10. A
−3 can't be factored. x−2 lim −3 can't be factored. x→2 x−2 A vertical as symptote occurs at x = 2. symptote A vertical as occurs at x = 2. 2x+1 1 lim e = e =e x→ 0 ex+1 e01 2x e lim = e =e x x→ 0 e e0 3 2 lim t − 4t = lim t t − 4 = t →−2 t →−2 t2+ 2 3t + 2 lim t − 4t = lim t t − 4 = t →−2 t →−2 t +2 t +2 (t + 2) (t − 2) t = lim t → −2 (t + 2) ( ) ( ) lim t t + 2 t − 2 = t → −2 (t + 2) lim t ( t − 2) = −2( −2 − 2) = Calculus −2( −4) = 8 lim x→2
( (
t →−2
) )
lim t ( t − 2) = −2( −2 − 2) = −2( −4) = 8
t →−2
limit from the left
x + 2 = 2+ 2 = 4 2+3 5 x+3 x + 2 = 2+ 2 = 4 2+3 5 x+3 3+ x = 3+1 = 2 − ln ( x) 2 − ln (1) 3+ x = 3+1 = 2 − ln ( x) 2 − ln (1) 2 sec ( θ ) = 2(1) = 2 −2 θ−2 0−2 2 sec ( θ ) = 2(1) = 2 −2 θ−2 0−2 x2 + x − 6 = x−2 x2 + x − 6 = x−2 (x + 3) (x − 2) = (x + x3)−(x2 − 2) = x−2 x+3=5
