Sec 5.1 and Sec 5.2 Exponential and logarithmic functions.

Any function of the type is an exponential function, with base , and(x)f = bx b

exponent ‘x’.

We must have . Or else something line can’t be defined. and b =b > 0 / 1 b0.5

If . In this case is the constant function 1., f(x)b = 1 = 1x = 1 (x)f

Domain of an exponential function: − , )( ∞ ∞

Range of an exponential function : 0, )( ∞

Every exponential function is continuous and differentiable every where.

Q. The picture below shows the graph of the functions . Match, , 5 , .1 , 0.2 , 0.51x 2x x 0 x x x

the graphs with their functions.

Soln. Orange -> 1x blue -> 2x pink -> 5x Turquoise -> 0.1x Green -> .20 x

Red -> .50 x

The graphs of all exponential functions have these properties.

1. The graphs lie above the x-axis, since . .bx > 0 =bx / 0

2. . So the graph of an exponential function has y-intercept 1.b0 = 1

3. They have the x-axis has a horizontal asymptote.

4. If , then and and is an increasing functionb > 1 limx → ∞

bx = ∞ .limx → −∞

bx = 0 bx

For example and limx → ∞

10x = ∞ .limx → −∞

10x = 0

5. If , then and and is a decreasing function.b < 1 limx → ∞

bx = 0 .limx → −∞

bx = ∞ bx

For example and limx → ∞

0.1x = 0 .limx → −∞

0.1x = ∞

Q. The picture below shows the graph of the functions . Match the functions, 0.32x x

with their graphs.

Soln. green -> . red -> .2x .30 x

Q. The picture below shows the graph of . Match the graph with their, 32x x

functions.

Soln. red -> , green -> .3x 2x

Q. The picture below shows the graph of . Match the graph with their, 0.20.5x x

functions. For the function , what is the exponent and what is the base.50 x

Soln. red -> , green -> .0.5x 0.2x

For the function , the exponent is x and the base is 0.5.50 x

Laws of Exponents:

Q. Simplify the given expressions:

5 5 /54 .(125)−34 . 4 .52.7 −1.3 2

= 42.7−1.3+3 2−3 = 44.4 −1 = 44.4

2(−12)33 271.4 2

= 3 31.4 6

(−1)(4) 33 3 = − 31.4+6−3 −6

( ) ] −[ 2−1 3 −2 = ( 1) (2 )−6 −1 −6 = 26

) ( ) 2 ) (3 2 ) 3 1/2)(1/3) /6( 116

−1/46427 −1/3 = ( −4 −1/4 3 6 −1/3 = 21−2 −1 = ( = 1

) ( ) 2 3 ) (3 2 ) 3 3 2( 827

−1/3 81256

−1/4 = ( 3 −3 −1/3 4 −8 −1/4 = 2−1 1 −1 2 = 2

) x y ) x y ) y(x y2n−2 2n

x y5n+1 −n1/3 = ( 2n−2−5n−1 2n+n 1/3 = ( −3n−3 3n 1/3 = x−n−1 n

(x−y)2(x+y)(x−y)

= (x+y)(x−y)1−2

= x−yx+y

2x )( −b/a −a/b = 2 x−a/b

4x )(− x ) 2x( −2 3 5 = − 1 3

Q. Solve the equation 792x = 2

Soln. First step is to bring both the sides of the equation to the same base.

.34x = 33

Once they are in the same base , then we must have the exponents must also be

equal.

792x = 2

⇒ 34x = 33

x⇒ 4 = 3 /4⇒ x = 3

Q. Solve )( 132

x−2 = 8x

Soln. )(2−5 x−2 = 23x

x 0 x⇒ − 5 + 1 = 3

x 0 0/8⇒ 8 = 1 ⇒ x = 1

Q. Solve 2.3 79x − 1 x + 2 = 0

Soln. 2.3 732x − 1 x + 2 = 0

at this stage if we take . then the equation looks like3 ) 2.3 7⇒ ( x 2 − 1 x + 2 = 0 3x = y

2y 7⇒ y2 − 1 + 2 = 0 y )(y )⇒ ( − 9 − 3 = 0

, y⇒ y = 9 = 3 ., or 3⇒ 3x = 9 x = 3

or x⇒ x = 2 = 1

Q. Solve .24x − 4 x + 4 = 0

Soln. . Set .222x − 4 x + 4 = 0 y = 2x y⇒ y2 − 4 + 4 = 0

. which has solution . which gives . which gives .y )⇒ ( − 2 2 = 0 y = 2 2x = 2 x = 1

Definition. There is a special number called the euler’s number denoted by e. It’s

value is approximately 2.72. It is also equal to the limit = (1e : limx → ∞

+ )x1 x

Q. Note , similarly . This is how we handle functions) .52−x = (21 x = 0 x 0.1) ) 0( −x = ( 1

0.1x = 1 x

of the type . The following picture shows the graph of the functionsb−x

. Match the graphs with their functions., 0.1) , , , .5 ,2−x ( −x e−x ex 0 −x 4−x

Soln. When written with positive exponents the functions are respectively

0.5) , 10 , (1/e) 0.37) , e 2.72) , , 0.25)( x x x ≃ ( x x ≃ ( x 2x ( x

Green -> 0.1)( −x

Pink -> ex Orange -> 0.5)( −x

Blue -> 4−x

Turquoise -> e−x

Red -> 2−x

Note: the graphs of are reflections of each other on the y-axis., eex −x

Q. If . Find f(3), if f(1)=5, f(2)=7(x) xef = A −kx

Soln. f(1)= ,eA −k = 5

.(2) Aef = 2 −2k = 7

So we we have to solve these equations simultaneously to find . If we divide,A k

the the two equations we get Ae−k

2Ae−2k = 75

0/7⇒ 2ek = 7

5 ⇒ ek = 1

e e 0/7A −k = 5 ⇒ A = 5 k = 5

(3) Ae (50/7)(e ) 150/7)(10/7) 150/7)(7/10)f = 3 −3k = 3 k −3 = ( −3 = ( 3 = 150.73

7.1000 = 10015.49 = 20

3.49

Important remark: To compare exponential functions ( for example to draw their

graphs on the same axes) we bring them to the common exponent. For example to

compare , we write them as For example to compare the functions44x, −x , (0.25) 4x x

we can write them as, , 0 , 0 , 1/2) , 1/4) , 2 , 3 2x 4x 1 x 1 −x ( x ( x −x −x

. So we have the ascending order of functions, , 0 , 0.1) , 0.5) , 0.25) , (0.5) , (0.33)2x 4x 1 x ( x ( x ( x x x

0 1/4) 01 −x < ( x = 4−x < 3−x < 2−x < 2x < 4x < 1 x

Sec 5.2 Logarithmic functions.

The function is the inverse of the function It is called the logarithmicog xl b .bx

function with base . By this we mean if . then we must have . Note b og xy = l b . by = x

we must have , and , =b > 0 b / 1 . x > 0

For example , since og 25l 5 = 2 552 = 2

since og 3 .5l 9 = 0 90.5 = 3 since og 9 /3l 27 = 2 27)( 2/3 = 9

since og (x x )l (x−1)2 − 2 + 1 = 2 x ) x( − 1 2 = x2 − 2 + 1

since og 3l (1/3) − 1 1/3)( −1 = 3

since og 1l anything = 0 anything)( 0 = 1 =1 since og (same thing)l anything anything) ame thing( 1 = s

since og xl x = 1 x1 = x og bl b

x = x

Q. Solve for x.og 0.5l x = − 1

Soln. .0.5) . So xx−1 = ( = 2−1 = 2

Domain of is since .og xy = l b 0, )( ∞ x = by > 0

Range of is (- since the domain of is .og xy = l b , )∞ ∞ bx − , )( ∞ ∞

Note if is the inverse of then,(x)f (x)g

, gf ° g = x ° f = x Domain of = range of (x)f (x)g

Range of Domain of (x)f = (x)g

The graph of is the reflection of the graph of along the line .(x)f (x)g y = x

The bases , and are very commonly used, and the logarithms to these0b = 1 b = e

bases have special names:

called the common logarithm of x, log 10 = 1og x og xl 10 = l

called the natural logarithm of x ln e = 1og x n xl e = l

Laws of Logarithms.

In the second law for logarithms, there should be a “ - “ instead of “ ? “

log(6)=log2+log3

ln(5/3)=ln(5)-ln(3)

log( ) 1/2)ln(7)√7 = (

og 1l 5 = 0 og 2000l 2000 = 1 og (x y ) log x log yl 3

2 3 = 2 3 + 3 3

og og (x ) og 2 og (x )l 2 2xx +12 = l 2

2 + 1 + l 2x = l 2

2 + 1 + x n ln x 0.5)ln (x ) xl exx2√x −12 = 2 + ( 2 − 1 −

og x log xl ( )b1 = − b

Proof: Let .og xy = l b

Then .by = x

Then )(b1 −y = x

So og x og xlb1 = − y = − l b

Q. If log 2= .3010, log 5 = .6990, log 3 = .4771

Find log 6, log 10, log 81, log 50, , log og 5l 0.1 √5

Soln. log 6= log (2.3)= log 2 + log 3 = .3010 + .4771 = .7781

Log 10 = log (2.5) = log 2 + log 5 = .3010 + .6990 = 1.000 (as it should be, since

the base is 10)

Log 81 = log ( 4 log 3 = 4 (.4771) = 1.9084)34 =

Log 50 = log ( )= 2 log 5 + log 2 = 1.3980 + .3010 = 1.6990.252

og 5 log 5 og 5 6990l 0.1 = − 10 = − l = − .

Log = (0.5)(.6990) = .34950.5)log 5√5 = (

Q. Write the expression as a logarithm of a single quantity.

n 3 ln x ln y ln z n 3 n n y n z nl + 21 + 3 − 3

1 = l + l √x + l 3 + l −1/3 = l z1/33( )y√x 3

n 2 ln(x ) ln(1 ) n[ ]l + 21 + 1 − 2 + √x = l

(1+ )√x 22(x+1)1/2

Q. Expand and simplify the expression.

og(x(x ) ) og x 1/2)log(x )l 2 + 1 −1/2 = l − ( 2 + 1

n ln x 1/2)ln x ln(1 )l x2

(1+x)√x 2 = 2 − ( − 2 + x

n(x(x )(x )) n x n (x ) n(x )l + 1 + 2 = l + l + 1 + l + 2

n n e n(e ) n(e )l exe +1x = l x − l x + 1 = x − l x + 1

n(xe ) n x n(e ) n xl −x2 = l + l −x2 = l − x2

The following figure shows the graph of , log (x)ex e

The graph of is got by reflecting the graph of about the line y=xog (x)l b bx

For example the graph of and areog xl 0.5 0.5)( x

The graph of has the following features.og xl b

1. The graph is only defined over x>0.

2. The x-intercept of is always 1. Since has the soln x=1og xl b og xl b = 0

3. The graph of , are symmetric about the line y=x.og xl b bx

4. The graph always has the y-axis as the vertical asymptote.

5. If b>1 is an increasing functionog xl b

6. If b<1 is a decreasing function.og xl b

7. Since . The graph of is the reflection about x-axis ofog x og xlb1 = − l b og xl (1/b)

the graph of .og xl b

For example the following diagram shows the graph of and . Note bothog xl 2 og xl 0.5

the graphs pass through (1,0) and have the y-axis as a vertical asymptote.

To sketch the graph of , we can draw the graph of , then reflect it aboutog xl b bx

the line y=x.

The following diagram compares how the graph of changes when we change theog xl b

base . In particular diagram shows the graphs ofb

. Note ifog x (red), log x (green), log x (turquoise), log x (blue), log x (orange), log x (pink)l 2 3 4 0.25 0.33 0.5

, is an increasing fn, if , is a decreasing fn.b > 1 og xl b b < 1 og xl b

Q. Sketch the graphs of and n(2x)y = l n( x)y = l 21

Soln. . ln2 > 0 .n(2x) n 2 n xl = l + l

. ln(0.5) < 0 .n( x) n( ) n xl 21 = l 2

1 + l

The graphs and are the red curve and the blue curven(2x)y = l n( x)y = l 21

respectively.

, , for any . blog xb = x og (b )l bx = x , =b > 0 b / 1

In particular , ln(e ) eln x = x x = x

For example , log (5 )5log 25 = 2 54 = 4

The above relationships help in solving the following equations.

Q. Solve e2 x+2 = 5

Soln. .5ex+2 = 2

n(e ) n(2.5)⇒ l x+2 = l

n(2.5)⇒ x + 2 = l

n(2.5)⇒ x = l − 2

Remark if we have in the exponent , and we want to bring it down, we can takex

logarithm to the appropriate base of both sides of the equation.

Q. Solve for t002001+3e−0.3t = 1

Soln. e1 + 3 −0.3t = 2 /3⇒ e−0.3t = 1

.3t ln(1/3) n(3)⇒ − 0 = = − l

ln(3)/0.3⇒ t =

Q. Solve for x2x = e3x

Soln. n(2 ) xl x = 3

ln(2) x⇒ x = 3

(ln(2) )⇒ x − 3 = 0 0⇒ x =

Q. Solve the equation: ln x 5 + 3 = 0

Soln. n x /5l = − 3

⇒ eln x = e−3/5

⇒ x = e−3/5

If we want to remove or , raise both sides of the equation to the exponentog (.)l b nl

of an appropriate base.

Q. If , write I in terms of D, .og( )D = l II0

I0

Soln. Since it involves a , which we want to remove, we raise both sides toog ( )l 10

the base 10.

og( )D = l II0

0 0⇒ 1 D = 1 log( )II0

0⇒ 1 D = II0

0 (I )⇒ I = 1 D0