Exponential_logarithm_1.pdf

Exponential & Logarithmic functions Exponential functions

y = f(x) = b

for b > 0 and b

¹

1

b = base x = exponent

Domain: all possible values of x: the set of real numbers Range: all possible values of y or f(x): y > 0

Horizontal asymptote: y = 0 (or) x-axis

It is one-to-one function. It passes both vertical line and horizontal line tests. Its inverse is also a function.

n

m _b

b =

ó m = n

x

x _c

b =

ó b = c

> b

ó m > n

Example 1: y = f(x) = 2x, b = 2 > 0

x y = 2x x y = 2x

0 20 = 1 à y-intercept (0,1) 0 20 = 1

-1

5 . 0 2 1

2-1 _{= =} 1 2

1 _{= 2}

-5

0.03125 32

1 2

1

2-5 _{= 5 = =} 2 2

2 _{= 4}

-10

= =

=

1024 1 2

1

2-10 _{10 0.0009765625} 10 2

10 _{= 1024}

-50

= =

=

-842624 1125899906

1 2

1

2 50 _{50 almost 0 but > 0} 50 2

50 _{= 1125899906842624}

x à -¥, y à zero from above

x-axis (y = 0) is the horizontal asymptote

x à +¥, y à +¥

For 0 < b < 1

, b

> b

ó m < n

Example 2: y = f(x) = (0.5)x, b = 0.5, 0 < b < 1

x y = (0.5)x x y = (0.5)x

0 (0.5)0 = 1 à y-intercept (0,1) 0 (0.5)0 = 1 -1 2 5 . 0 1 ) 5 . 0 ( 1 ) 5 . 0

( -1 _{= 1 = =} 1 (0.5)1 = 0.5

-2 4 25 . 0 1 ) 5 . 0 ( 1 ) 5 . 0

( -2 _{= 2 = =} 2 (0.5)2 = 0.25

-5 32 0.03125 1 ) 5 . 0 ( 1 ) 5 . 0

( -5 _{= 5 = =} 5 (0.5)5 = 0.03125

-10 1024 25 0.00097656 1 ) 5 . 0 ( 1 ) 5 . 0

( -10 _{= 10 = =} 10 (0.5)10 = 0.0009765625

x à -¥, then y à +¥ x à +¥, then y à zero from above

x-axis is asymptote Example 3

Graph the function g(x) = (-3)32x +1₊1

Think about the simplest exponential function with the same base 3, f(x) = 3x g(x) = (-3)32x +1₊1

g(x) = -3 f(2x + 1) + 1 g(x) = -3 f[2(x + 0.5)] + 1 g(x) is transformation of f(x)

- vertical reflection (reflection about x-axis) - vertical stretch by the factor of 3

- horizontal stretch by the factor of ½ - horizontal shift to the left by 0.5 unit - vertical shift up by 1 unit

3x = f(x)

32x+1 = f(2x +1)

Transformation table (x, y)

y = f(x) y = -f(x) (x, y) reflect on x-axis

(x, y) y = -3 f(x) vertical stretch x3

(x, y) y = -3 f(2x) hori stretch x1/2

(x, y) y = -3 f[2(x+0.5)] Horiz shift 0.5 left

-3 1/27 -3 -1/27 -3 -1/9 -1.5 -1/9 -2 -1/9

-1 1/3 -1 -1/3 -1 -1 -1/2 -1 -1 -1

0 1 0 -1 0 -3 0 -3 -0.5 -3

1 3 1 -3 1 -9 1/2 -9 0 -9

(x, y)

y = -3 f[2(x+0.5)] + 1 vertical shift 1 up

-2 8/9

-1 0

-0.5 -2

0 -8

f(x) to g(x) transformation

(-3, 1/27) à (-2, 8/9)

(-1, 1/3) à (-1, 0)

(0, 1) à (-0.5, -2)

(1, 3) à (0, -8)

Example 4:

Solve the inequality:

243 1 32x+1_³

Inequality in one variable “x”. Check with the calculator, 243 = 35 243

1 32x+1_³

5 1 2x

3 1 3 + ³

5 5 1 2x

3 1 3 + ³

5 1

2x

3 1

3 ÷

ø ö ç è æ ³ +

( )

2x ₃

3 + _³ -5 1

2x ₃

3 + ³

-Base 3 > 1

Therefore if ₃2x+1_³3-5 _then

2x + 1 ³ -5 2x ³ -6

x ³ -3 [Divide both sides by positive 2]

** If we can not make both sides (of inequality or equation) to have the same base, take logarithm both sides.

Example 5:

Solve the inequality: (0.5)2x – y > 2 and verify your answer with 3 sample answers. Inequality in two variables: x and y. The answer will be some portion(s) of x-y plane. (0.5)2x - y > 2

(0.5)2x - y - 2> 0 [subtract 2 from both sides] (0.5)2x - 2> y [add y on both sides]

y < (0.5)2x – 2 [rewrite to have y on left side]

Test with 3 points in the shaded area: (-3, 2) => x = -3 and y = 2

(0.5)2x – 2 = (0.5)2(-3) – 2 = (0.5)-6 – 2 = 62 > y (-2, -2) => x = -2 and y = -2

(0.5)2x – 2 = (0.5)2(-2) – 2 = (0.5)-4 – 2 = 14 > y (2, -3) => x = 2 and y = -3

(0.5)2x – 2 = (0.5)2(2) – 2 = (0.5)4 – 2 = -1.9375 > y

Example 6:

Write an exponential function of the graph that passes through (0, 6.4) and (3, 100)

We only have two information that two points (0, 6.4) and (3, 100) are on the graph. Therefore, we choose the function with only two unknowns:

y = a bx [a and b are to be found] (0, 6.4) => x = 0 and y = 6.4

y = a bx 6.4 = a b0 6.4 = a

(3, 100) => x = 3 and y = 100 y = a bx

100 = (6.4) b3 100/6.4 = b3

3

b 4 . 6 100

=

b 4 . 6 100

3 =

2.5 = b

Example 7

Write an exponential inequality with a solution of x £ 2. We know that if b > 1

bx£ b2 if and only if x £ 2

So one answer is: 3x£ 32 [b = 3 > 1]

Exponential and Logarithm

Inverse of

_is

x

by ₌

_{is the same as}

b > 0 and b

¹

1

For b > 1

x

b y =

Domain: x belongs to the set of real numbers Range: y > 0

x-axis (y = 0) is horizontal asymptote It is increasing function

bm > bn ó m > n

It is inverse of y =log_bx (or) _{x =b}y

It is reflection of y =log_bx (or) _{x =b}y _{about the line y = x}

x log

y = _b (or) _{x =b}y

Domain: x > 0

Range: y belongs to the set of real numbers y-axis (x = 0) is vertical asymptote

It is increasing function n

log m

log_b > _b ó m > n It is inverse of _{y =b}x

For b > 1

bm > bn ó m > n n log m

For 0 < b < 1

x

b y =

Domain, range, asymptote are same as above Decreasing function

bm > bn ó m < n x

log

y = _b (or) _{x =b}y

Domain, range, asymptote are same as above Decreasing function

n log m

log_b > _b ó m < n

Properties of Logarithm M

log_b is real only if M > 0, b > 0 and b ¹ 1 M

log ß Base is not written - - assume base b = 10 M

ln = log_eM where e = about 2.71828 is a mathematical constant M

log_b = log_bN ó M = N b

log_b = 1 log 10 = log₁₀10 =1 ln e = log_ee =1 1

log_b = 0

N log M log (MN)

log_b = _b + _b

N log -M log N

M

log_b ÷= _{b b}

ø ö ç è æ

M log n M

log n _b

b =

X blogbX =

For 0 < b < 1 bm > bn ó m < n

n log m

b log

A log A log

c c

b = [base change formula, c > 0 and c ¹ 1]

Example 8 (refer to example 6):

In example 6 we have y = (6.4) (2.5)x as one answer for the question.

Write an exponential function of another graph that also passes through (0, 6.4) and (3, 100). This time we use a different model y = bx + c where we have two unknowns: b and c.

y = bx + c must pass the two given points (0, 6.4) and (3, 100) (0, 6.4) => x = 0 and y = 6.4

y = bx + c

6.4 = b(0) + c = 1 + c 5.4 = c

(3, 100) => x = 3 and y = 100 y = bx + c

100 = b3 + 5.4 94.6 = b3

b 94.6

3 ₌

4.5565 = b

Answer: y = 4.5565x + 5.4

y = f(x) = (6.4) (2.5)x passes the two given points (0, 6.4) & (3, 100) y = g(x) = 4.5565x + 5.4 also passes these two points.

f(x) and g(x) are two different graphs and functions. They intersect each other at these two points. f(x) is based on the exponential model: y = a bx

g(x) is based on the exponential model: y = bx + x

Example 9:

Graph the function: y = f(x) = log (2x – 1) + 1 Domain

2x – 1 > 0 [For log (M) to be real, M must be greater than zero] 2x > 1

x > 0.5

Base is not mentioned so base b = 10

Think about the simplest logarithmic function with base 10: y = g(x) = log (x) log (x) = g(x)

log (2x – 1) = g(2x – 1)

log (2x – 1) + 1 = g(2x – 1) + 1 = g[2(x - 0.5)] + 1 f(x) = g[2(x - 0.5)] + 1

Graph g(x) = log (x) and do the transformation to get f(x). Transformation from g(x) to f(x)

- horizontal stretch by the factor of 1/2 - horizontal shift of 0.5 unit right - vertical shift of 1 unit up

Transformation from g(x) to f(x) (x, y)

y = g(x)

(x, y) y = g(2x) Horiz Stretch x 0.5

(x, y) y = g[2(x – 0.5)] Horiz shift 0.5 right

(x, y)

y = g[2(x – 0.5)] + 1 Vertical shift 1 up

0.1 -1 0.05 -1 0.55 -1 0.55 0

1 0 0.5 0 1 0 1 1

10 1 5 1 5.5 1 5.5 2

Vertical asymptote (y-axis, x = 0) of g(x) will not be moved by - horizontal stretch and

- vertical shift

x = 0.5 is vertical asymptote of y = f(x)

Domain for f(x): x > 0.5

Range for f(x): y is any real number x-intercept = (0.55, 0)

y-intercept = none