Math Lesson 13.2.1 - The Definition of Exponential Equations

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Welcome to our Math lesson on The Definition of Exponential Equations, this is the first lesson of our suite of math lessons covering the topic of Exponential and Logarithmic Equations, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

The Definition of Exponential Equations

As the name suggests, exponential equations are equations that have the variable(s) in the exponent. For example,

are all examples of exponential equations, as all variables are in the exponent.

Exponential equations have different methods of solution, depending on the complexity of the parts involved. In the following paragraphs of this tutorial, we will explain how to solve exponential equations from the easiest to the most challenging.

1. Exponential equations with a single exponential term

The simplest case of exponential equations involves situations where there is a single exponential term on one side and the result on the other side of the equation, which is a power of the base number that contains the variable in the exponent. The general form of such equations is

For example,

belongs to this type of exponential equation, as there is a single exponential term 2x on the left side and a single number on the right that is a power of 2.

The first thing to check when trying to solve such equations is to try writing the result as an exponential number with the same base as the term that contains the variable. If you can do this, then you can forget the common base and focus only on the exponential part, where you obtain a type of equation (usually linear) as those we have discussed earlier in this course.

For example, in the exponential equation above, we can express the right side as 2³ instead of 8. In this way, we obtain

Since the base is the same on both sides, so there must the exponents be as well. In this way, we obtain

Example 1

Find the value of x in the exponential equations below.

1. 3^x = 243
2. 2^4x = 64

Solution 1

1. Given that 243 = 3⁵, we have
   3^x = 243
   3^x = 3⁵
   Now, the bases are the same. Hence, the exponents must be the same as well. Thus,
   x = 5
2. Given that 64 = 2⁶, we have
   2^4x = 64
   2^4x = 2⁶
   Now, the bases are the same. Hence, the exponents must be the same as well. Thus,
   4x = 6
   x = 6/4
   x = 3/2

2. Exponential equations with more than one exponential term where all terms can be written at the same base

If the exponential equation contains more than one exponential part, the solution procedure may require some kind of factorisation. However, if it is possible two write all terms at the same base, this is not a problem. Let's consider a pair of examples to illustrate this point.

Example 2

Calculate the value of the variable x in the following exponential equations.

1. 5 + 2^{3x - 1} = 3²
2. 63 - 3^x = 2 ∙ 3^{x + 1}

Solution 2

1. At first glance, this equation seems to have two exponential parts but one of them contains just numbers, so we can complete the operations to simplify the number part and write it as a power of 2. We have
   5 + 2^{3x - 1} = 3²
   5 + 2^{3x - 1} = 9
   2^{3x - 1} = 9 - 5
   2^{3x - 1} = 4
   2^{3x - 1} = 2²
   After having expressed both sides as powers of the same base (here the base is 2), we can focus on the exponents only. Thus,
   3x - 1 = 2
   3x = 2 + 1
   3x = 3
   x = 3/3
   x = 1
2. We can express 3x + 1 as 3 · 3x. Thus,
   63 - 3^x = 2 ∙ 3^{x + 1}
   63 - 3^x = 2 ∙ 3 ∙ 3^x
   63 - 3^x = 6 ∙ 3^x
   63 = 6 ∙ 3^x + 3^x
   Factoring 3x on the right side yields
   63 = 3^x (6 + 1)
   63 = 7 ∙ 3^x
   3^x = 63/7
   3^x = 9
   3^x = 3²
   Thus, since the bases are already the same on both sides, we obtain
   x = 2

3. Exponential equations where the variable appears in more than one term

As you see, the exponential equations are becoming more and more challenging. However, if you can express every term to the same base, it is still possible to solve such exponential equations through the standard methods (i.e. without involving logarithms). Let's consider a pair of examples to clarify this point.

Example 3

Solve the following exponential equations.

1. 2³x = 160 - 4 ∙ 8^x
2. 40 - 5^4x/25^x = 3

Solution 3

1. First, let's separate all terms containing the variable from the constants. Thus,
   2^3x = 160 - 4 ∙ 8^x
   2³x + 4 ∙ 8^x = 160
   Now, let's express all terms containing the variable at the same power. Thus,
   2^3x + 4 ∙ (2³ )^x = 160
   2^3x + 4 ∙ 2³x = 160
   Factoring the common term 2^3x yields
   (1 + 4) ∙ 2^3x = 160
   5 ∙ 2^3x = 160
   2^3x = 160/5
   2^3x = 32
   2^3x = 2⁵
   3x = 5
   x = 5/3
2. Again, first let's separate all terms containing the variable from the constants. Thus,
   40 - 5^{2x + 1}/25^x = 3
   40 - 5^{2x + 1} = 3 ∙ 25^x
   40 = 3 ∙ 25^x + 5^{2x + 1}
   Now, we express the terms containing the variable to the same base. Thus,
   40 = 3 ∙ (5² )^x + 5^{2x + 1}
   40 = 3 ∙ 5²x + 5^{2x + 1}
   Moreover, we can express 5^{2x + 1} as 5 · 5^2x. Hence,
   40 = 3 ∙ 5^2x + 5 ∙ 5^2x
   40 = (3 + 5) ∙ 5^2x
   40 = 8 ∙ 5^2x
   40/8/ = 5^2x
   5 = 5^2x
   5¹ = 5^2x
   1 = 2x
   x = 1/2

4. Solving exponential equations where the terms cannot turn to a common base

Now things become more challenging, as if the terms are impossible to write at the same base, we must use the logarithmic approach to solve these exponential equations. Obviously, the base of the logarithm on both sides must be equal to the argument of the term that contains the variable. Only in this way, we can eliminate the exponent from the equation. The simplest case would be when there is a single exponential term and a constant on the other side of the equation but the logarithmic method can be extended further to include a wide range of exponential equations. Let's consider a few examples in this regard.

Example 4

Solve the following exponential equations.

1. 3^x = 12
2. 2^5x/7 = 4^3x - 3
3. 2^{x - 1}/9 = 8^2x

Solution 4

1. We have
   3^x = 12
   log₃ 3^x = log₃ 12
   x = log₃ 12
   x = log₃ (3 ∙ 4)
   x = log₃ 3 + log₃ 4
   x = 1 + log₃ 4
   The exercise is considered as solved in this form, as log₃ 4 is a number that has a certain value. Indeed, looking at the calculator provided by us, we obtain the value of 1.26 when written to two decimal places. Hence, we obtain
   x = 1 + 1.26
   = 2.26
   Remark! We can also find the value of log₃ 4 by expressing it as below.
   log₃ 4 = log 4/log 3
   = 0.602/0.477
   = 1.26
2. 2^5x/7 = 4^3x - 3
   7 ∙ 2^5x/7 = 7 ∙ (4^3x - 3)
   2^5x = 7 ∙ 4^3x - 7 ∙ 3
   2^5x = 7 ∙ (2² )^3x - 7 ∙ 3
   2^5x = 7 ∙ 2^{2 ∙ 3x} - 21
   2^5x = 7 ∙ 2^6x - 21
   2^5x = 7 ∙ 2 ∙ 2^5x - 21
   2^5x = 14 ∙ 2^5x - 21
   21 = 14 ∙ 2⁵x - 2^5x
   21 = (14 - 1) ∙ 2^5x
   21 = 13 ∙ 2^5x
   2^5x = 21/13
   (2⁵)^x = 21/13
   32^x = 21/13
   log₃₂ 32^x = log₃₂ 21/13
   x = log₃₂ 21 - log₃₂ 13
   x = log 21/log 32 - log 13/log 32
   = log 21 - log 13/log 32
   = 1.32 - 1.11/1.51
   = 0.21/1.51
   = 0.14
3. We have
   2^{x - 1}/9 = 8^2x
   2^{x - 1} = 9 ∙ 8^2x
   2^{x - 1} = 9 ∙ (2³ )^2x
   2^{x - 1} = 9 ∙ 2^{3 ∙ 2x}
   2^{x - 1} = 9 ∙ 2⁶x
   2^x ∙ 2^-1 = 9 ∙ 2^6x
   1/2 ∙ 2^x = 9 ∙ 2^6x
   2^x = 9 ∙ 2 ∙ 2^6x
   2^x = 18 ∙ 2^6x
   1/18 = 2⁶x/2^x
   1/18 = 2^{6x - x}
   1/18 = 2^5x
   log₂ 1/18 = log₂ 2⁵x
   5x = log₂ 18^-1
   5x = -log₂ 18
   5x = -log 18/log 2
   x = -1/5 ∙ log 18/log 2
   x = -1/5 ∙ log 18/log 2
   x = -1/5 ∙ 1.255/0.301
   x = -1.255/1.505
   = 0.83

5. Exponential equations where the variable is written to the second power

Sometimes, the variable of an exponential equation is in the second power. In these cases we must solve a quadratic equation when turning all terms at the same base is impossible; otherwise, we use the logarithmic approach. Let's consider a pair of examples to clarify this point.

Example 5

Solve the following exponential equations.

1. 3^{x2 + 1} = 9^x
2. 15 ∙ 2⁵x = 3 ∙ 2^x2

Solution 5

1. First, let's write all terms at the same base. Thus,
   3^{x2 + 1} = 9^x
   3^{x2 + 1} = (3² )^x
   3^{x2 + 1} = 3²x
   Now, we focus only on the exponential part. Thus,
   x² + 1 = 2x
   x²-2x + 1 = 0
   This quadratic equation has a single root, as it can be written in the form
   (x-1)² = 0
   Hence, the solution is
   x = 1
2. It is impossible to write all terms of this equation to the same base. This means we have to use the logarithmic approach for solving it. We have
   15 ∙ 2⁵x = 3 ∙ 2^x2
   15/3 = 2^x2/2⁵x
   5 = 2^{x2 - 5x}
   log₂ 5 = log₂ 2^{x2 - 5x}
   log₂ 5 = (x² - 5x) ∙ log₂ 2
   log₂ 5 = (x² - 5x) ∙ 1
   Since log₂ 5 ≈ 0.4, we obtain
   x² - 5x = 0.4
   x² - 5x - 0.4 = 0
   We can multiply all terms by 5 to remove decimals. Thus, we obtain
   x² - 5x - 0.4 = 0
   5x² - 25x - 2 = 0
   We have a = 5, b = -25 and c = -2. Thus,
   x₁ = -b - √b² - 4ac/2a
   = -(-25) - √(-25)² - 4 ∙ 5 ∙ (-2)/2 ∙ 5
   = 25 - √625 + 40/10
   = 25 - √665/10
   = 25 - 25.8/10
   = -0.08
   and
   x₁ = -b - √b² - 4ac/2a
   = 25 + 25.8/10
   = 5.08

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