Solving Exponential Equations Using Logarithms and Factoring

Exponential equations are a fundamental concept in mathematics, often appearing in various fields such as physics, engineering, and finance. This article will demonstrate how to solve a specific exponential equation step-by-step, using both logarithms and factoring methods. We will explore each step in detail to ensure a comprehensive understanding of the underlying principles.

Introduction to Exponential Equations

An exponential equation is an equation where the variable appears in the exponent. For example, in the equation (3^{3frac{2x-1}{x}} 3^{2frac{2x 1}{2}}), both sides are expressed with the same base, which makes it easier to solve using logarithms or by equating the exponents directly.

Solving the Given Exponential Equation

Consider the equation 27^{frac{2x-1}{x}} 9^{1 frac{2x}{2}}.

Step 1: Simplify the Bases

First, rewrite 27 and 9 in terms of base 3:

27 3^3 and 9 3^2.

Therefore, the equation becomes:

3^{3frac{2x-1}{x}} 3^{2 cdot 1 x} 3^{2x 1}.

Step 2: Equate the Exponents

Since the bases are the same, we can equate the exponents:

frac{6x-3}{x} 2x 1.

Step 3: Solve the Equation

Multiply both sides by x to clear the denominator:

6x - 3 2x^2 x.

Bring all terms to one side to form a quadratic equation:

2x^2 - 5x - 3 0.

Step 4: Factor the Quadratic Equation

Factor the quadratic equation:

(2x 1)(x - 3) 0.

Step 5: Find the Roots

Solve for x by setting each factor equal to zero:

2x 1 0 Rightarrow x -frac{1}{2}

x - 3 0 Rightarrow x 3

Verification

Check the solutions by substituting them back into the original equation:

For (x 1.5):

27^{frac{2 cdot 1.5 - 1}{1.5}} 27^{frac{3 - 1}{1.5}} 27^{frac{2}{1.5}} 27^{frac{4}{3}} 27^1 cdot 3 27 cdot 3 81

9^{1 frac{2 cdot 1.5}{2}} 9^{1 frac{3}{2}} 9^{frac{5}{2}} 9^2 cdot 3 81

For (x 1):

27^{frac{2 cdot 1 - 1}{1}} 27^{frac{2 - 1}{1}} 27^1 27

9^{1 frac{2 cdot 1}{2}} 9^{1 1} 9^2 27

Both solutions are verified.

Conclusion

Thus, the solutions to the equation (27^{frac{2x-1}{x}} 9^{1 frac{2x}{2}}) are (x 1.5) and (x 1).

Understanding and solving exponential equations is crucial for many real-world applications. This step-by-step approach provides a clear and concise method to solve such equations, making it easier to tackle more complex problems in mathematics.