Solving Exponential Equations: The 3 proven Methods and Techniques in solving exponential equations

What are Exponential Equations?

An exponential equation is an equation where the variable is in the exponent. For example, 2^x = 8 is an exponential equation, where x is the variable. Exponential equations can also have variables in the base, such as 3^x = y, where x is the variable and y is a constant.

Exponential equations can be classified into two types: linear and nonlinear. Linear exponential equations are those in which the variable appears only in the exponent. They can be solved using logarithms or by simplifying the equation to a linear form. Nonlinear exponential equations are those in which the variable appears in both the exponent and the base. These equations cannot be solved using traditional methods, and numerical methods are needed to approximate the solution.

Methods for Solving Exponential Equations:

Simplification Method:

Another method for solving linear exponential equations is the simplification method. This method involves simplifying the exponential equation into a linear form, which can then be solved using basic algebraic techniques.

Consider the equation 3^x = 81. To solve this equation using the simplification method, we can express both sides of the equation in terms of the same base, say base 3, as follows:

3^x = 3^4

Since both sides of the equation have the same base, we can equate the exponents and solve for x as follows:

x = 4

Thus, the solution to the exponential equation 3^x = 81 is x = 4.

Numerical Methods:

Nonlinear exponential equations cannot be solved using traditional algebraic methods. In such cases, numerical methods such as the bisection method, the Newton-Raphson method, or the fixed-point iteration method can be used to approximate the solution.

Consider the equation 2^x = x + 1. This equation cannot be solved using traditional algebraic methods. However, we can use the bisection method to approximate the solution.

The bisection method involves dividing the interval in which the solution lies into smaller intervals and testing each interval until the solution is found. To use the bisection method for the equation 2^x = x + 1, we can begin by plotting the two functions f(x) = 2^x and g(x) = x + 1 on the same graph. The point where these two curves intersect is the solution to the equation.

We can start by choosing an interval that contains the solution, such as [0, 1]. We then evaluate the function at the midpoint of the interval and check if it has the same sign as either endpoint. If it does not, we have found the solution. Otherwise, we divide the interval in half and repeat the process until we converge to the solution.

Using the bisection method, we can approximate the solution to the equation 2^x = x + 1 as follows:

• Starting interval: [0, 1]
• Midpoint: 0.5
• Evaluate function at midpoint: f(0.5) = 2^0.5 – 1 = 0.414
• Function has different signs at endpoints
• New interval: [0, 0.5]
• Midpoint: 0.25
• Evaluate function at midpoint: f(0.25) = 2^0.25 – 1 = -0.354
• Function has different signs at endpoints
• New interval: [0.25, 0.5]
• Midpoint: 0.375
• Evaluate function at midpoint: f(0.375) = 2^0.375 – 1 = 0.028
• Function has different signs at endpoints
• New interval: [0.375, 0.5]
• Midpoint: 0.4375
• Evaluate function at midpoint: f(0.4375) = 2^0.4375 – 1 = -0.165
• Function has different signs at endpoints
• New interval: [0.4375, 0.5]
• Midpoint: 0.46875
• Evaluate function at midpoint: f(0.46875) = 2^0.46875 – 1 = -0.069
• Function has different signs at endpoints
• New interval: [0.46875, 0.5]
• Midpoint: 0.484375
• Evaluate function at midpoint: f(0.484375) = 2^0.484375 – 1 = -0.020
• Function has different signs at endpoints
• New interval: [0.484375, 0.5]
• Midpoint: 0.4921875
• Evaluate function at midpoint: f(0.4921875) = 2^0.4921875 – 1 = 0.004
• Function has different signs at endpoints
• New interval: [0.484375, 0.4921875]
• Midpoint: 0.48828125
• Evaluate function at midpoint: f(0.48828125) = 2^0.48828125 – 1 = -0.008
• Function has different signs at endpoints
• New interval: [0.48828125, 0.4921875]
• Midpoint: 0.490234375
• Evaluate function at midpoint: f(0.490234375) = 2^0.490234375 – 1 = -0.002
• Function has different signs at endpoints
• New interval: [0.490234375, 0.4921875]
• Midpoint: 0.4912109375
• Evaluate function at midpoint: f(0.4912109375) = 2^0.4912109375 – 1 = 0.001
• Function has different signs at endpoints
• New interval: [0.490234375, 0.4912109375]

• Midpoint: 0.49072265625
• Evaluate function at midpoint: f(0.49072265625) = 2^0.49072265625 – 1 = -0.0005
• Function has different signs at endpoints
• New interval: [0.49072265625, 0.4912109375]
• Midpoint: 0.490966796875
• Evaluate function at midpoint: f(0.490966796875) = 2^0.490966796875 – 1 = 0.0003
• Function has different signs at endpoints
• New interval: [0.49072265625, 0.490966796875]
• Midpoint: 0.4908447265625
• Evaluate function at midpoint: f(0.4908447265625) = 2^0.4908447265625 – 1 = -0.0001
• Function has different signs at endpoints
• New interval: [0.4908447265625, 0.490966796875]
• Midpoint: 0.49090576171875
• Evaluate function at midpoint: f(0.49090576171875) = 2^0.49090576171875 – 1 = 0.0001
• Function has different signs at endpoints
• New interval: [0.4908447265625, 0.49090576171875]
• Midpoint: 0.490875244140625
• Evaluate function at midpoint: f(0.490875244140625) = 2^0.490875244140625 – 1 = 0
• Function has same sign at both endpoints
• Solution: 0.490875244140625

As we can see, the bisection method requires a lot of iterations to converge to the solution. However, it is guaranteed to converge as long as the function is continuous and changes sign over the interval.

Logarithmic Method:

Another way to solve exponential equations is to use logarithms. Recall that the logarithm is the inverse of the exponential function, i.e., log_a(b) = x if and only if a^x = b.

Suppose we have an equation of the form a^x = b, where a is a positive real number. We can take the logarithm of both sides with respect to base a to get:

log_a(a^x) = log_a(b)

x log_a(a) = log_a(b)

x = log_a(b)

This gives us the solution to the equation. For example, to solve the equation 2^x = 8, we can take the logarithm of both sides with respect to base 2 to get:

log_2(2^x) = log_2(8)

x log_2(2) = 3

x = 3

Therefore, the solution to the equation is x = 3.

Note that we can use logarithms with any base, as long as we use the same base on both sides of the equation. Also, if the equation is of the form a^x = a^y, we can take the logarithm of both sides with respect to base a to get:

log_a(a^x) = log_a(a^y)

x log_a(a) = y log_a(a)

x = y

This tells us that if a^x = a^y, then x = y. In other words, if the bases are equal, then the exponents must be equal.