📈 Graph of Exponential and Logarithmic Functions

Sahabat Smaridasa — In this article, we will discuss the graph of exponential and logarithmic functions. The graph of an exponential function is a monotonic graph, either monotonically increasing or decreasing. However, in this article on Graph of Exponential and Logarithmic Functions, we will only discuss the exponential function graph. As for the graph of logarithmic functions, we have previously shared it in an article entitled "logarithmic functions". Please visit that article to learn about logarithmic function graphs.

Drawing the Graph of an Exponential Function is not very difficult. The simplest form of an exponential function is $ f(x) = a^x $. Please also read the material on "exponential functions" to make it easier to study and create/draw the graph of an exponential function. The main factor that determines the shape of the exponential function graph is the value of $ a $, which is called the base (please read: General Form of Exponents or Powers). If $ a > 1 $, the graph is generally monotonically increasing, and if $ 0 < a < 1 $, the graph is monotonically decreasing.

📊 Graph of Exponential Function $ f(x) = a^x $

Graph of $ f(x) = a^x $

For $ a > 1 $: The graph intersects the Y-axis at $ y = 1 $ and is monotonically increasing.
For $ 0 < a < 1 $: The graph intersects the Y-axis at $ y = 1 $ and is monotonically decreasing.

Note: We can take several points $(x, y)$ that satisfy the exponential function by substituting chosen values of $ x $ first, then after substitution we will get the value of $ y $. These points will help us in drawing the graph more easily.

Example 1

Draw the graph of the following exponential functions:
a) $ f(x) = 2^x $    b) $ f(x) = 5^x $    c) $ f(x) = 9^x $
d) $ f(x) = \left(\frac{1}{2}\right)^x $    e) $ f(x) = \left(\frac{1}{5}\right)^x $    f) $ f(x) = \left(\frac{1}{9}\right)^x $

Solution:

• For $ f(x) = 2^x, \, f(x) = 5^x, \, f(x) = 9^x $, the base is greater than 1, so the graph is monotonically increasing as shown below.

• For $ f(x) = \left(\frac{1}{2}\right)^x, \, f(x) = \left(\frac{1}{5}\right)^x, \, f(x) = \left(\frac{1}{9}\right)^x $, the base is between 0 and 1, so the graph is monotonically decreasing as shown below.

Note: The graph of $ f(x) = \left(\frac{1}{a}\right)^x $ can be obtained by reflecting the graph of $ f(x) = a^x $ across the Y-axis, and vice versa.

📊 Graph of Exponential Function $ f(x) = b \cdot a^x $

Graph of $ f(x) = b \cdot a^x $

For $ a > 1 $: The graph intersects the Y-axis at $ y = b $ and is monotonically increasing.
For $ 0 < a < 1 $: The graph intersects the Y-axis at $ y = b $ and is monotonically decreasing.

Example 2

Draw the graph of the exponential functions $ f(x) = 2 \cdot 5^x $ and $ f(x) = 2 \cdot \left(\frac{1}{5}\right)^x $.

Solution:

📊 Graph of Exponential Function $ f(x) = b \cdot a^x + c $

Graph of $ f(x) = b \cdot a^x + c $

For $ a > 1 $: The graph intersects the Y-axis at $ y = b + c $ and is monotonically increasing.
For $ 0 < a < 1 $: The graph intersects the Y-axis at $ y = b + c $ and is monotonically decreasing.

Example 3

Draw the graph of the following exponential functions:
a) $ f(x) = 2 \cdot 3^x + 1 $    b) $ f(x) = 2 \cdot 3^x - 3 $
c) $ f(x) = 2 \cdot \left(\frac{1}{3}\right)^x + 1 $    d) $ f(x) = 2 \cdot \left(\frac{1}{3}\right)^x - 3 $

Solution:

• For (a) and (c): $ b = 2 $ and $ c = 1 $, so the Y-intercept is $ y = 2 + 1 = 3 $.
• For (b) and (d): $ b = 2 $ and $ c = -3 $, so the Y-intercept is $ y = 2 - 3 = -1 $.

Graphs (a) and (b) are monotonically increasing:

Graphs (c) and (d) are monotonically decreasing:

📊 Graph of Negative Exponential Functions

Negative Exponential Functions

The graphs of $ f(x) = -a^x $, $ f(x) = -b \cdot a^x $, and $ f(x) = -(b \cdot a^x + c) $ are obtained by reflecting the graphs of $ f(x) = a^x $, $ f(x) = b \cdot a^x $, and $ f(x) = b \cdot a^x + c $ across the X-axis.

Example 4

Draw the graph of the following exponential functions:
a) $ f(x) = -2 \cdot 3^x $    b) $ f(x) = -2 \cdot 3^x + 3 $

Solution:

• a) The graph of $ f(x) = -2 \cdot 3^x $ is obtained by reflecting the graph of $ f(x) = 2 \cdot 3^x $ across the X-axis.

• b) The graph of $ f(x) = -2 \cdot 3^x + 3 = -(2 \cdot 3^x - 3) $ is obtained by reflecting the graph of $ f(x) = 2 \cdot 3^x - 3 $ across the X-axis.

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Thus concludes our discussion on the Graph of Exponential and Logarithmic Functions along with examples. Next, please also read other related materials such as determining the exponential function from its graph. May this material be beneficial. Thank you.

💡 Tip: The best way to master graphing exponential functions is through continuous practice. Try drawing various exponential function graphs with different bases and constants.

📐 Properties of Exponents

Sahabat Smaridasa — To make it easier to work with exponential forms, we must know the properties of exponents that will be used in solving problems. These properties of exponents are very important and play the most crucial role in exponentiation. Therefore, if you want to be proficient and comfortable in mastering and solving exponent problems, we must first master the properties well and correctly.

There are many properties of exponents or exponentiation that we must memorize. However, it is important to remember that memorizing alone is not enough; we must also know how to use each property properly. If we can remember and use all these properties well, then we can say we have successfully learned them. Don't worry, in this article we have prepared the properties along with examples for each.

Problems that are directly related to properties of exponents or exponentiation always appear every year, both in national exams and in university entrance tests. This means that by mastering the properties of exponents well, at least one problem is guaranteed to be solvable. To deepen the use of exponent properties, please read and work on the collection of exponent problems available.

🔢 Properties of Exponents Based on the Exponent

The properties of exponents can be classified based on the type of exponent: positive integers, zero, negative integers, and fractions.

(i) Positive integer exponents ( \(m, n \in \mathbb{Z}^+\) )

\[ \begin{aligned} 1)&\; a^{m} \cdot a^{n} = a^{m+n} \\ 2)&\; \frac{a^{m}}{a^{n}} = a^{m-n} \\ 3)&\; (a^{m})^{n} = a^{m \cdot n} \\ 4)&\; (ab)^{m} = a^{m} \cdot b^{m} \\ 5)&\; \left(\frac{a}{b}\right)^{m} = \frac{a^{m}}{b^{m}} \end{aligned} \]

(ii) Zero exponent

\[ a^{0} = 1, \quad \text{with the condition } a \neq 0 \]

(iii) Negative integer exponent ( \(n\) positive)

\[ a^{-n} = \frac{1}{a^{n}} \quad \text{or} \quad \frac{1}{a^{-n}} = a^{n} \]

(iv) Fractional exponents

\[ \begin{aligned} 1)&\; a^{\frac{1}{n}} = \sqrt[n]{a} \\ 2)&\; a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m} \end{aligned} \]

Below are some examples for the properties of exponents mentioned above.

Example 1

Determine the value of \(2^{3} \cdot 2^{4}\) and \(5^{2} \cdot 5^{2}\).

Solution: \[ \begin{aligned} &2^{3} \cdot 2^{4} = 2^{3+4} = 2^{7} = 128 \\ &5^{2} \cdot 5^{2} = 5^{2+2} = 5^{4} = 625 \end{aligned} \]

Example 2

Determine the value of \(\dfrac{5^{4}}{5^{3}}\) and \(\dfrac{7^{8}}{7^{6}}\).

Solution: \[ \begin{aligned} &\frac{5^{4}}{5^{3}} = 5^{4-3} = 5^{1} = 5 \\ &\frac{7^{8}}{7^{6}} = 7^{8-6} = 7^{2} = 49 \end{aligned} \]

Example 3

Determine the value of \((3^{2})^{2}\).

Solution: \[ (3^{2})^{2} = 3^{2 \cdot 2} = 3^{4} = 81 \]

Example 4

Determine the value of \((2 \cdot 3)^{2}\) and \(5^{3} \cdot 2^{3}\).

Solution: \[ \begin{aligned} &(2 \cdot 3)^{2} = 2^{2} \cdot 3^{2} = 4 \cdot 9 = 36 \\ &5^{3} \cdot 2^{3} = (5 \cdot 2)^{3} = 10^{3} = 1000 \end{aligned} \]

Example 5

Determine the value of \(\left(\dfrac{2}{5}\right)^{2}\) and \(\dfrac{6^{3}}{2^{3}}\).

Solution: \[ \begin{aligned} &\left(\frac{2}{5}\right)^{2} = \frac{2^{2}}{5^{2}} = \frac{4}{25} \\ &\frac{6^{3}}{2^{3}} = \left(\frac{6}{2}\right)^{3} = 3^{3} = 27 \end{aligned} \]

Example 6

Determine the value of \(5^{0}\) and \(\left(-\frac{1}{3}\right)^{0}\).

Solution: \[ \begin{aligned} &5^{0} = 1 \\ &\left(-\frac{1}{3}\right)^{0} = 1 \end{aligned} \]

Example 7

Determine the value of \(2^{-3}\), \(\dfrac{1}{2^{-2}}\), and \(\dfrac{3}{2^{-4}}\).

Solution: \[ \begin{aligned} &2^{-3} = \frac{1}{2^{3}} = \frac{1}{8} \\ &\frac{1}{2^{-2}} = 2^{2} = 4 \\ &\frac{3}{2^{-4}} = 3 \cdot 2^{4} = 3 \cdot 16 = 48 \end{aligned} \]

Example 8

Determine the value of \(4^{\frac{1}{2}}\) and \(3^{\frac{1}{5}}\).

Solution: \[ \begin{aligned} &4^{\frac{1}{2}} = \sqrt{4} = 2 \\ &3^{\frac{1}{5}} = \sqrt[5]{3} \end{aligned} \]

Example 9

Determine the value of \(3^{\frac{2}{3}}\) and \(16^{\frac{3}{4}}\).

Solution: \[ \begin{aligned} &3^{\frac{2}{3}} = \sqrt[3]{3^{2}} = \sqrt[3]{9} \\ &16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^{3} = 2^{3} = 8 \\ &\text{Alternative: } 16^{\frac{3}{4}} = (2^{4})^{\frac{3}{4}} = 2^{3} = 8 \end{aligned} \]

Note: For \(n = 2\), the number 2 does not need to be written in the fractional exponent, as it is commonly known as the square root form.
\[ \begin{aligned} &2^{\frac{1}{2}} \text{ can be written as } \sqrt{2} \\ &\text{Method: } 2^{\frac{1}{2}} = \sqrt[2]{2} = \sqrt{2} \\ &3^{\frac{5}{2}} \text{ can be written as } \sqrt{3^{5}} \\ &\text{Method: } 3^{\frac{5}{2}} = \sqrt[2]{3^{5}} = \sqrt{3^{5}} \end{aligned} \]

⚠️ Exponent properties with negative base

\[ (-a)^{n} = \begin{cases} a^{n}, & \text{for } n \text{ even} \\ -(a^{n}), & \text{for } n \text{ odd} \end{cases} \]

Example 10

Determine the value of \((-2)^{4}\) and \((-2)^{5}\).

Solution: \[ \begin{aligned} &(-2)^{4} = 2^{4} = 16 \quad (\text{even exponent, positive result}) \\ &(-2)^{5} = -(2^{5}) = -32 \quad (\text{odd exponent, negative result}) \end{aligned} \]

We hope the examples above help us understand the properties of exponents, which I think are quite numerous. The best way to remember all properties of exponents is to continuously practice with exponent problems. We are confident that with practice, you will naturally remember and master them, including how to use each property.

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💡 Tip: The best way to master all properties is through continuous practice. Work through many exponent problems until you naturally remember each rule and its application.

Algebraic Function Limits - Practice Problems

Sahabat Smaridasa - Mathematical Concepts

In this article, we will work on Algebraic Function Limit Practice Problems to reinforce the material on "Solving Algebraic Function Limits". Previously, we have also discussed "Definition of Function Limits" and "Properties of Function Limits". Below are practice problems that we can solve to strengthen our understanding.

The problems come in various forms, but most are rational expressions that lead to the indeterminate form \(\frac{0}{0}\). Therefore, we need to process them further using methods such as:

• Factorization – factoring and canceling common terms.
• Rationalization (Multiplying by the Conjugate) – especially for limits involving square roots.
• L'Hôpital's Rule (Derivatives) – another powerful method.

💡 With consistent practice, even challenging problems involving roots will become easier to handle.

Practice Problems

Understanding the Concept of Function Limit

Sahabat Smaridasa - Mathematical Foundation

In everyday life, we often hear words like "almost" or "approaching". For example, "Messi almost scored a goal", "the speed of the motorcycle is approaching 110 km/h", and so on. The words "almost" or "approaching" in mathematics are called limits. In this article, we will study the Definition of Function Limits. The function limits discussed here refer to "algebraic function limits" and "trigonometric function limits" which will be covered in other articles. In mathematics, a limit represents the approximate value of a variable as it approaches a real number. The following is the limit notation.

Definition of Function Limit

Let \( f \) be a function \( f : \mathbb{R} \to \mathbb{R} \) and let \( L \) and \( a \) be real numbers.

\[ \lim_{x \to a} f(x) = L \]

if and only if \( f(x) \) approaches \( L \) for all \( x \) approaching \( a \).

How to read limit notation: \( \lim_{x \to a} f(x) = L \) is read as "the limit of function \( f(x) \) as \( x \) approaches \( a \) equals \( L \)".

Methods for Evaluating Limits

To determine the limit value of a function, there are several methods:

1. Numerical Method (tabulation)
2. Direct Substitution
3. Factorization
4. Multiplying by the Conjugate
5. Using Derivatives (L'Hôpital's rule)

In this article on the definition of limits, we will only use the numerical method. The numerical method is a way of calculating limits by substituting values from the left and right sides and listing results in a table. However, this method is less efficient because building a table takes time.

📘 Example 1 :

Find the limit of the function \( f(x) = x + 1 \) as \( x \) approaches \( 2 \).

Solution:

1. The problem can be written as: \(\displaystyle \lim_{x \to 2} (x + 1) = \ldots\)
2. Using the numerical method, we choose values of \( x \) approaching 2 from the left and right, then substitute them into \( (x + 1) \). The results are shown in the following table.

Table of values for \( f(x) = x + 1 \) as \( x \) approaches 2

\(x\)	1.9	1.99	1.999	1.9999	→ 2 ←	2.0001	2.001	2.01	2.1
\(f(x)=x+1\)	2.9	2.99	2.999	2.9999	?	3.0001	3.001	3.01	3.1

Gambar 1: Ilustrasi nilai fungsi \( f(x) = x + 1 \) saat x mendekati 2 (pendekatan numerik)

From the table and illustration above, we see that from the left side of 2, the function values approach 2.9999. From the right side of 2, the function values approach 3.0001. This means that the limit of \( f(x) = x + 1 \) as \( x \) approaches 2 is 3. Hence, \(\displaystyle \lim_{x \to 2} (x + 1) = 3\).

Condition for a Function to Have a Limit at a Point

A limit is said to exist if the left-hand limit and the right-hand limit are equal. The left-hand limit is the value the function approaches from the left, denoted \(\displaystyle \lim_{x \to a^-} f(x)\). The right-hand limit is the value the function approaches from the right, denoted \(\displaystyle \lim_{x \to a^+} f(x)\).

That is, if \(\displaystyle \lim_{x \to a^-} f(x) = L\) and \(\displaystyle \lim_{x \to a^+} f(x) = L\), then \(\displaystyle \lim_{x \to a} f(x) = L\).

Description of whether a limit exists for a function \( f(x) \) as \( x \) approaches \( c \).

Gambar 2: Ilustrasi keberadaan limit berdasarkan limit kiri dan kanan (Figure A, B, C, D)

Based on the graphical description above:

• Figure A : The limit exists because the function approaches the same value from both sides (continuous).
• Figure B : The limit does not exist because the left-hand limit ≠ right-hand limit (jump discontinuity).
• Figure C : The limit exists because the left-hand limit = right-hand limit, despite a hole at the point.
• Figure D : The limit does not exist because the left-hand limit ≠ right-hand limit (different asymptotic behavior).

📘 Example 2 :

Does the following function have a limit as \( x \) approaches \( 1 \)?

\[ f(x) = \begin{cases} x^2, & \text{if } x \le 1 \\ x + 1, & \text{if } x > 1 \end{cases} \]

Solution:

• If \( x \le 1 \), then \( f(x) = x^2 \).
• If \( x > 1 \), then \( f(x) = x + 1 \).

Table of values approaching 1 from the left and right:

\(x\)00.50.70.90.990.999→ 1 ←1.0011.011.11.51.7 \(f(x)\)00.250.490.810.98010.998001?2.0012.012.12.52.7

Analysis of left-hand and right-hand limits:
Left-hand limit: from the left approaching 1, the function values approach \(0.998001 \approx 1\), so \(\displaystyle \lim_{x \to 1^-} f(x) = 1\).
Right-hand limit: from the right approaching 1, the function values approach \(2.001 \approx 2\), so \(\displaystyle \lim_{x \to 1^+} f(x) = 2\).
Because the left-hand limit and right-hand limit are not equal, the function \( f(x) = \begin{cases} x^2 & x \le 1 \\ x+1 & x > 1 \end{cases} \) as \( x \) approaches 1 does NOT have a limit.

Gambar 3: Grafik fungsi \( f(x) = \begin{cases} x^2 & x \le 1 \\ x+1 & x > 1 \end{cases} \) di sekitar x = 1 (loncatan)

⚠️ Conclusion: The function \( f(x) = \begin{cases} x^2, & x \le 1 \\ x+1, & x > 1 \end{cases} \) as \( x \) approaches 1 does NOT have a limit.

Properties of Function Limits

Sahabat Smaridasa - Mathematical Concepts

To make it easier to determine the limit value of a function, we need what are called properties of function limits. These properties are theorems used to solve limits of functions. There are various methods to compute limits, one of which is direct substitution, which we will use in this article. You may also read the material on "Definition of Function Limits".

Evaluating Limits by Direct Substitution

The substitution method means directly substituting the value \(x\) into the function \(f(x)\). For example: \(\displaystyle \lim_{x \to a} f(x) = f(a)\).

📘 Example 1 :

Determine the limit values of the following forms:

a) \(\displaystyle \lim_{x \to 2} (2x + 1)\)     b) \(\displaystyle \lim_{x \to -1} (2x - 1)\)

Solution:

a) \(\displaystyle \lim_{x \to 2} (2x + 1) = 2(2) + 1 = 4 + 1 = 5\).
Hence, \(\displaystyle \lim_{x \to 2} (2x + 1) = 5\).

b) \(\displaystyle \lim_{x \to -1} (2x - 1) = 2(-1) - 1 = -2 - 1 = -3\).
Hence, \(\displaystyle \lim_{x \to -1} (2x - 1) = -3\).

Properties of Function Limits

The following are the properties of function limits (assuming all individual limits exist):

• i) \(\displaystyle \lim_{x \to a} k = k\), where \(k\) is a constant.
• ii) \(\displaystyle \lim_{x \to a} [k \cdot f(x)] = k \cdot \lim_{x \to a} f(x)\).
• iii) \(\displaystyle \lim_{x \to a} [f(x) \pm g(x)] = \lim_{x \to a} f(x) \pm \lim_{x \to a} g(x)\).
• iv) \(\displaystyle \lim_{x \to a} [f(x) \cdot g(x)] = \left( \lim_{x \to a} f(x) \right) \left( \lim_{x \to a} g(x) \right)\).
• v) \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{ \lim\limits_{x \to a} f(x) }{ \lim\limits_{x \to a} g(x) }\), provided \(\lim\limits_{x \to a} g(x) \neq 0\).
• vi) \(\displaystyle \lim_{x \to a} [f(x)]^n = \left[ \lim_{x \to a} f(x) \right]^n\) for any integer \(n\).
• vii) \(\displaystyle \lim_{x \to a} \sqrt{f(x)} = \sqrt{ \lim_{x \to a} f(x) }\), provided the limit inside the root is non-negative (for even roots).

📘 Example 2 :

Determine the following limits using the properties of limits:

a) \(\displaystyle \lim_{x \to 2} 5\)    b) \(\displaystyle \lim_{x \to 3} 2x^3\)    c) \(\displaystyle \lim_{x \to 1} (x^2 + x)\)    d) \(\displaystyle \lim_{x \to -1} (x^2 - 3x)\)

e) \(\displaystyle \lim_{x \to -2} (x^3 \cdot x^2)\)    f) \(\displaystyle \lim_{x \to 3} \frac{x+1}{x^2-1}\)    g) \(\displaystyle \lim_{x \to 2} (2x^2 + 3)^9\)    h) \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1}\)

Solution (using limit properties):

a) \(\displaystyle \lim_{x \to 2} 5 = 5\).

b) \(\displaystyle \lim_{x \to 3} 2x^3 = 2 \cdot \lim_{x \to 3} x^3 = 2 \cdot 3^3 = 2 \cdot 27 = 54\).

c) \(\displaystyle \lim_{x \to 1} (x^2 + x) = \lim_{x \to 1} x^2 + \lim_{x \to 1} x = 1^2 + 1 = 1 + 1 = 2\).

d) \(\displaystyle \lim_{x \to -1} (x^2 - 3x) = \lim_{x \to -1} x^2 - 3\lim_{x \to -1} x = (-1)^2 - 3(-1) = 1 + 3 = 4\).

e) \(\displaystyle \lim_{x \to -2} (x^3 \cdot x^2) = \left( \lim_{x \to -2} x^3 \right) \left( \lim_{x \to -2} x^2 \right) = (-2)^3 \cdot (-2)^2 = (-8) \cdot 4 = -32\).

f) \(\displaystyle \lim_{x \to 3} \frac{x+1}{x^2-1} = \frac{ \lim_{x \to 3} (x+1) }{ \lim_{x \to 3} (x^2-1) } = \frac{3+1}{3^2 - 1} = \frac{4}{9-1} = \frac{4}{8} = \frac{1}{2}\).

g) \(\displaystyle \lim_{x \to 2} (2x^2 + 3)^9 = \left( \lim_{x \to 2} (2x^2 + 3) \right)^9 = \left( 2 \cdot 2^2 + 3 \right)^9 = (2 \cdot 4 + 3)^9 = (8+3)^9 = 11^9\).

h) \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1} = \sqrt{ \lim_{x \to 3} (x^2 - 1) } = \sqrt{3^2 - 1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}\).

💡 Note: All solutions above use the limit properties listed. Substitution works because these functions are continuous at the given points.

📘 Example 3 :

Evaluate \(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1}\) using properties (alternate illustration).

Solution:

\(\displaystyle \lim_{x \to 3} \sqrt{x^2 - 1} = \sqrt{ \lim_{x \to 3} (x^2 - 1) } = \sqrt{3^2 - 1} = \sqrt{9-1} = \sqrt{8} = 2\sqrt{2}\).

Thus, the limit equals \(2\sqrt{2}\).

Table of Contents: INTEGRAL

1. General Definition of Integral
2. Difference Between Definite and Indefinite Integrals
3. Indefinite Integrals of Algebraic Functions
4. Determining Curve Equations Using Indefinite Integrals
5. Indefinite Integral of Trigonometric Functions
6. Algebraic Substitution Integration Technique
7. Integration by Parts Technique
8. Trigonometric Integration
9. Integral of Exponential Functions
10. Advanced Integration Techniques
11. Fundamental Theorem of Calculus
12. Proof of Fundamental Theorem of Calculus
13. Computation and Properties of Definite Integrals
14. Riemann Sums in Integrals
15. Calculating Area Using Integrals
16. Quick Methods for Calculating Area Between Curves
17. Volume of Solids of Revolution Using Integrals
18. Determining Arc Length Using Integrals
19. Determining Integrals of Absolute Value Functions