Understanding Logarithmic and Exponential Functions Graphs

$x$	$1013 e^{- 0.000128 x}$	$y = 1013 e^{- 0.000128 x}$
$0$	$1013 e^{- 0.000128 (0)}$	$1013$
$1000$	$1013 e^{- 0.000128 (1000)}$	$\approx 891$
$2000$	$1013 e^{- 0.000128 (2000)}$	$\approx 784$
$3000$	$1013 e^{- 0.000128 (3000)}$	$\approx 690$
$4000$	$1013 e^{- 0.000128 (4000)}$	$\approx 607$
$5000$	$1013 e^{- 0.000128 (5000)}$	$\approx 534$
$6000$	$1013 e^{- 0.000128 (6000)}$	$\approx 470$
$7000$	$1013 e^{- 0.000128 (7000)}$	$\approx 414$
$8000$	$1013 e^{- 0.000128 (8000)}$	$\approx 364$
$9000$	$1013 e^{- 0.000128 (9000)}$	$\approx 320$
$10000$	$1013 e^{- 0.000128 (10000)}$	$\approx 282$

x

1013 e^{- 0.000128 x}

y = 1013 e^{- 0.000128 x}

0

1013 e^{- 0.000128 (0)}

1013

1000

1013 e^{- 0.000128 (1000)}

\approx 891

2000

1013 e^{- 0.000128 (2000)}

\approx 784

3000

1013 e^{- 0.000128 (3000)}

\approx 690

4000

1013 e^{- 0.000128 (4000)}

\approx 607

5000

1013 e^{- 0.000128 (5000)}

\approx 534

6000

1013 e^{- 0.000128 (6000)}

\approx 470

7000

1013 e^{- 0.000128 (7000)}

\approx 414

8000

1013 e^{- 0.000128 (8000)}

\approx 364

9000

1013 e^{- 0.000128 (9000)}

\approx 320

10000

1013 e^{- 0.000128 (10000)}

\approx 282

$x$	$2 lo g (x - 0.5)$	$c (x) = 2 lo g (x - 0.5)$
$1$	$2 lo g (1 - 0.5)$	$\approx - 0.6$
$2$	$2 lo g (2 - 0.5)$	$\approx 0.35$
$3$	$2 lo g (3 - 0.5)$	$\approx 0.8$
$4$	$2 lo g (4 - 0.5)$	$\approx 1.1$
$5$	$2 lo g (5 - 0.5)$	$\approx 1.3$

x

2 lo g (x - 0.5)

c (x) = 2 lo g (x - 0.5)

1

2 lo g (1 - 0.5)

\approx - 0.6

2

2 lo g (2 - 0.5)

\approx 0.35

3

2 lo g (3 - 0.5)

\approx 0.8

4

2 lo g (4 - 0.5)

\approx 1.1

5

2 lo g (5 - 0.5)

\approx 1.3

$x$	$2 lo g (x - 1)$	$f (x)$
$1.5$	$2 lo g (1.5 - 1)$	$\approx - 0.6$
$5$	$2 lo g (5 - 1)$	$\approx 1.2$
$10$	$2 lo g (10 - 1)$	$\approx 1.9$
$12$	$2 lo g (12 - 1)$	$\approx 2.1$
$18$	$2 lo g (18 - 1)$	$\approx 2.5$

x

2 lo g (x - 1)

f (x)

1.5

2 lo g (1.5 - 1)

\approx - 0.6

5

2 lo g (5 - 1)

\approx 1.2

10

2 lo g (10 - 1)

\approx 1.9

12

2 lo g (12 - 1)

\approx 2.1

18

2 lo g (18 - 1)

\approx 2.5

$x$	$1 0^{\frac{1}{2} x} + 1$	$f^{- 1} (x)$
$- 2$	$1 0^{\frac{1}{2} (- 2)} + 1$	$1.1$
$- 1$	$1 0^{\frac{1}{2} (- 1)} + 1$	$\approx 1.3$
$0$	$1 0^{\frac{1}{2} (0)} + 1$	$2$
$1$	$1 0^{\frac{1}{2} (1)} + 1$	$\approx 4.2$
$2$	$1 0^{\frac{1}{2} (2)} + 1$	$11$

x

1 0^{\frac{1}{2} x} + 1

f^{- 1} (x)

- 2

1 0^{\frac{1}{2} (- 2)} + 1

1.1

- 1

1 0^{\frac{1}{2} (- 1)} + 1

\approx 1.3

0

1 0^{\frac{1}{2} (0)} + 1

2

1

1 0^{\frac{1}{2} (1)} + 1

\approx 4.2

2

1 0^{\frac{1}{2} (2)} + 1

11

$x$	$\frac{1}{5} e^{3 x}$	$g (x)$
$- 1$	$\frac{1}{5} e^{3 (- 1)}$	$\approx 0.01$
$- 0.5$	$\frac{1}{5} e^{3 (- 0.5)}$	$\approx 0.04$
$0$	$\frac{1}{5} e^{3 (0)}$	$0.2$
$0.5$	$\frac{1}{5} e^{3 (0.5)}$	$\approx 0.9$
$1$	$\frac{1}{5} e^{3 (1)}$	$\approx 4$

x

\frac{1}{5} e^{3 x}

g (x)

- 1

\frac{1}{5} e^{3 (- 1)}

\approx 0.01

- 0.5

\frac{1}{5} e^{3 (- 0.5)}

\approx 0.04

0

\frac{1}{5} e^{3 (0)}

0.2

0.5

\frac{1}{5} e^{3 (0.5)}

\approx 0.9

1

\frac{1}{5} e^{3 (1)}

\approx 4

$x$	$\frac{1}{3} ln 5 x$	$g^{- 1} (x)$
$0.1$	$\frac{1}{3} ln 5 (0.5)$	$\approx - 0.2$
$1$	$\frac{1}{3} ln 5 (1)$	$\approx 0.5$
$2$	$\frac{1}{3} ln 5 (2)$	$\approx 0.8$
$3$	$\frac{1}{3} ln 5 (3)$	$\approx 0.9$
$4$	$\frac{1}{3} ln 5 (4)$	$\approx 1$

x

\frac{1}{3} ln 5 x

g^{- 1} (x)

0.1

\frac{1}{3} ln 5 (0.5)

\approx - 0.2

1

\frac{1}{3} ln 5 (1)

\approx 0.5

2

\frac{1}{3} ln 5 (2)

\approx 0.8

3

\frac{1}{3} ln 5 (3)

\approx 0.9

4

\frac{1}{3} ln 5 (4)

\approx 1

	Definition of the First Function	Substitute the Second Function	Simplify
$g (g^{- 1} (x))$	$\frac{1}{5} e^{3 g^{- 1} (x)}$	$\frac{1}{5} e^{3 (\frac{1}{3} l n 5 x)}$	$x ✓$
$g^{- 1} (g (x))$	$\frac{1}{3} ln 5 g (x)$	$\frac{1}{3} ln 5 (\frac{1}{5} e^{3 x})$	$x ✓$

Definition of the First Function

Substitute the Second Function

Simplify

g (g^{- 1} (x))

\frac{1}{5} e^{3 g^{- 1} (x)}

\frac{1}{5} e^{3 (\frac{1}{3} l n 5 x)}

x ✓

g^{- 1} (g (x))

\frac{1}{3} ln 5 g (x)

\frac{1}{3} ln 5 (\frac{1}{5} e^{3 x})

x ✓

An asymptote of a graph is an imaginary line that the graph gets close to as $x$ tends towards positive or negative infinity or a particular number. Find the asymptotes of the graphs.

Write the equation of the asymptote.

Write the equation of the asymptote.

This given graph is one of an exponential function. The domain of an exponential function is the set of all real numbers, which means that the graph cannot have a vertical asymptote.

Exponential Function	y = e^x -1
Domain	All real numbers
Asymptote	Horizontal

If we pay close attention, we can see that when x tends to negative infinity, the graph approaches the horizontal line y=- 1.

Therefore, y=- 1 is a horizontal asymptote.

This graph is the graph of a logarithmic function. Since the argument of a logarithm must be positive, the domain of this type of function is not the set of all real numbers. This means that the graph has a vertical asymptote.

Logarithmic Function	y = log(x+2)
Domain	Not all real numbers
Asymptote	Vertical

If we pay close attention, we can see that x approaches - 2, but its value never is - 2.

Therefore, x=- 2 is a vertical asymptote.

Considering the equations and graphs below. Order the functions from least to greatest average rate of change over the interval that goes from $1$ to $10 .$

Let's find the rate of change for each function in the interval 1≤ x ≤ 10. We will consider each function one at a time.

Option A

To find the rate of change for this logarithmic function in the interval from 1 to 10, we need to recall that the rate of change is the ratio of the change in y to the change in x. Rate of Change=y_2-y_1/x_2-x_1 Next, we will substitute x_1=1, x_2=10, y_1=log_6 1, and y_2=log_6 10 into the formula. Recall that the calculator only calculates logarithms with base 10 or base e — common and natural logarithms. Therefore, we will use the Change of Base Formula to convert the logarithms to common logarithms.

For the function in option A, the rate of change in the interval 1≤ x ≤ 10 is about 0.143.

Option B

We will find the rate of change for this function in the interval from 1 to 10 by following the same procedure as in option A. Let's do it!

Rate of change for y=log_(35) x in the interval 1≤ x ≤ 10
Formula	Substitute values	Evaluate
y_2-y_1/x_2-x_1	log_(35) 10- log_(35)1/10- 1	≈ - 0.501

Therefore, for the function in option B, the rate of change in the interval 1≤ x ≤ 10 is about - 0.501.

Option C

By using the graph, we can find the y-coordinates for x=1 and x=2.

We see in the graph that when x=1, the value of y is 1. Also, when x=10, the value of y is - 2. Let's use these values to find the rate of change in the interval 1≤ x≤ 10.

Rate of change for the graph in the interval 1≤ x ≤ 10
Formula	Substitute values	Evaluate
y_2-y_1/x_2-x_1	- 2- 1/10- 1	≈ - 0.333

Therefore, for the function in option C, the rate of change in the interval 1≤ x ≤ 10 is - 0.333.

Option D

Finally, to find the rate of change for the curve shown in option D, we will follow the same procedure as in option C. Let's do it!

We see that when x=1, the value of y is 0. Also, when x=10, the value of y is 8. We can use these values to find the rate of change for this curve in the interval that goes from 1 to 10.

Rate of change for the graph in the interval 1≤ x ≤ 10
Formula	Substitute values	Evaluate
y_2-y_1/x_2-x_1	8- 0/10- 1	≈ 0.889

Therefore, for the function in option D, the rate of change in the interval 1≤ x ≤ 10 is about 0.889.

Conclusion

Now that we have the four rates of change in the interval 1≤ x ≤ 10, we can write the options in order from least to greatest.

Option	B	C	A	D
Rate of change	≈ - 0.501	≈ - 0.333	≈ 0.143	≈ 0.889

Consider the function.

y = lo g_{b} x

Here,

b

is a positive real number different to

1 .

Determine if the statement below is always, sometimes, or never true.

The domain of the function contains $x = 0$ in its domain.

Let's use the definition of a logarithm to rewrite the equation. Definition:& c=log_b a ⇔ b^c= a Equation:& y=log_b x ⇔ b^y= x We found that x is equal to b^y. It is given that b is a positive real number. We know that any power of a positive number is always a positive number. Therefore, for any value of y, the value of b^y, or x, will always be positive. x > 0 Since the domain is the set of x-values and x>0, the domain of the function never contains 0.

	11 Theory slides
	16 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions

Graphing Exponential and Logarithmic Functions

Catch-Up and Review

Proof

Answer

Hint

Solution

Answer

Hint

Solution

Proof

logb​bx=x

blogb​x=x

Answer

Hint

Solution

Graphing p(x)=lnx

Graphing c(x)=2log(x−0.5)

Hint

Solution

Checking Our Answer

Exponential Graphs

Logarithmic Graphs

Option A

Option B

Option C

Option D

Conclusion

Graphing Exponential and Logarithmic Functions

Recommended exercises

$lo g_{b} b^{x} = x$

$b^{l o g_{b} x} = x$

Graphing $p (x) = ln x$

Graphing $c (x) = 2 lo g (x - 0.5)$