Solving Exponential Equations A Detailed Guide To 6^(2x-3) = 6^(-2x+1)

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In the realm of mathematics, solving exponential equations is a fundamental skill, crucial for tackling various problems in algebra, calculus, and real-world applications. Exponential equations, characterized by variables appearing in the exponents, demand a unique set of strategies for finding solutions. This comprehensive guide delves into the step-by-step solution of the equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1}, elucidating the underlying principles and techniques involved. Whether you're a student grappling with algebra or a math enthusiast seeking to deepen your understanding, this exploration will provide valuable insights and enhance your problem-solving prowess.

Understanding Exponential Equations

Before diving into the specifics of the equation at hand, it's essential to grasp the nature of exponential equations. At their core, these equations involve a constant base raised to a variable exponent. The goal is to determine the value(s) of the variable that satisfy the equation. The key to solving such equations lies in the properties of exponents and the ability to manipulate expressions to isolate the variable. When you're dealing with exponential equations, it’s important to remember the basic rules of exponents such as the product rule, quotient rule, power rule, and the zero and negative exponent rules. These rules provide the foundational tools to simplify and solve exponential equations. Moreover, understanding the exponential function and its properties is critical. The exponential function, f(x)=axf(x) = a^x, where aa is a constant base, is a cornerstone in various fields including finance, biology, and physics. Recognizing that the exponential function is one-to-one for a positive base aa not equal to 1, allows us to equate the exponents when the bases are the same, which is a key technique in solving exponential equations.

The Equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1} A Detailed Solution

The equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1} presents a classic example of an exponential equation where the bases are the same. This similarity is a crucial advantage, allowing us to directly equate the exponents and simplify the problem. The process begins by recognizing the equality of the bases. Since both sides of the equation have the same base (6), we can confidently set the exponents equal to each other. This step is rooted in the fundamental property of exponential functions: if am=ana^m = a^n, then m=nm = n, provided aa is a positive number not equal to 1. This property is the cornerstone of solving exponential equations with identical bases. By equating the exponents, we transform the exponential equation into a linear equation, which is far simpler to solve. This transformation is a critical step, streamlining the solution process and reducing the complexity of the problem. Linear equations are easier to manipulate and solve, which is why this approach is so effective. The equation now becomes 2xβˆ’3=βˆ’2x+12x - 3 = -2x + 1. This is a straightforward linear equation that can be solved using basic algebraic techniques. We proceed by isolating the variable terms on one side and the constant terms on the other. This involves adding 2x2x to both sides of the equation to eliminate the βˆ’2x-2x term on the right side, and adding 3 to both sides to eliminate the -3 term on the left side. These steps maintain the balance of the equation while moving terms to their appropriate sides. After these operations, the equation simplifies to 4x=44x = 4. The final step in solving for xx is to divide both sides of the equation by the coefficient of xx, which is 4. This isolates xx on one side, giving us the solution. Dividing both sides by 4, we get x=1x = 1. This is the solution to the original exponential equation.

Step-by-Step Solution

  1. Recognize the common base: In the equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1}, both sides have the base 6.
  2. Equate the exponents: Since the bases are equal, we can set the exponents equal to each other: 2xβˆ’3=βˆ’2x+12x - 3 = -2x + 1.
  3. Isolate the variable terms: Add 2x2x to both sides of the equation: 2x+2xβˆ’3=βˆ’2x+2x+12x + 2x - 3 = -2x + 2x + 1, which simplifies to 4xβˆ’3=14x - 3 = 1.
  4. Isolate the constant terms: Add 3 to both sides of the equation: 4xβˆ’3+3=1+34x - 3 + 3 = 1 + 3, which simplifies to 4x=44x = 4.
  5. Solve for x: Divide both sides by 4: 4x/4=4/44x / 4 = 4 / 4, which gives x=1x = 1.

Verification of the Solution

To ensure the accuracy of the solution, it's crucial to verify it by substituting the value of xx back into the original equation. This process confirms that the solution satisfies the equation and that no errors were made during the solving process. Substituting x=1x = 1 into the original equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1}, we get: 62(1)βˆ’3=6βˆ’2(1)+16^{2(1)-3} = 6^{-2(1)+1}. Simplifying the exponents, we have 62βˆ’3=6βˆ’2+16^{2-3} = 6^{-2+1}, which further simplifies to 6βˆ’1=6βˆ’16^{-1} = 6^{-1}. This equality confirms that x=1x = 1 is indeed the correct solution to the equation. Verification is a vital step in problem-solving, especially in mathematics. It not only confirms the correctness of the answer but also enhances understanding of the problem and the solution process. By substituting the solution back into the original equation, we are essentially retracing our steps, ensuring that each step was logically sound and mathematically accurate. This practice builds confidence and reinforces the problem-solving skills.

Common Mistakes to Avoid

When solving exponential equations, several common mistakes can lead to incorrect solutions. Being aware of these pitfalls can help in avoiding them and improving accuracy. One common mistake is incorrectly applying the properties of exponents. For example, students may try to multiply the base and the exponent, which is a fundamental error. It's crucial to remember the correct rules for manipulating exponents. Another frequent mistake is failing to recognize the importance of having the same base on both sides of the equation. If the bases are different, the exponents cannot be directly equated. In such cases, other techniques such as logarithms may be necessary. A third common error is making algebraic mistakes when solving the resulting linear equation. Errors in adding, subtracting, multiplying, or dividing terms can lead to an incorrect value for xx. Therefore, it's essential to carefully check each step of the algebraic manipulation. Lastly, neglecting to verify the solution is a significant oversight. Even if the solution seems correct, it's possible that an error was made during the solving process. Verification provides a final check and ensures that the solution satisfies the original equation.

Alternative Methods for Solving Exponential Equations

While equating exponents is a straightforward method for solving exponential equations with the same base, other techniques are required when the bases are different or the equation is more complex. One such method involves using logarithms. Logarithms are the inverse of exponential functions, and they provide a powerful tool for solving equations where the variable is in the exponent. Logarithms can be used to bring the exponent down, making it easier to isolate the variable. Another technique is to rewrite the equation in terms of a common base. If the bases can be expressed as powers of the same number, the equation can be simplified by substituting these expressions. For example, if one base is 4 and the other is 8, both can be written as powers of 2. Additionally, some exponential equations can be solved by substitution. This involves replacing a complex expression with a single variable, simplifying the equation, and then solving for the new variable. Once the value of the new variable is found, it can be substituted back to find the value of the original variable. These alternative methods provide a broader toolkit for solving a wider range of exponential equations.

Conclusion

In summary, solving the exponential equation 62xβˆ’3=6βˆ’2x+16^{2x-3} = 6^{-2x+1} demonstrates a fundamental principle in algebra: when the bases are the same, we can equate the exponents. This straightforward approach simplifies the problem into a linear equation, which is then easily solved. The solution, x=1x = 1, is verified by substituting it back into the original equation, confirming its correctness. Furthermore, understanding common mistakes and alternative methods for solving exponential equations broadens our problem-solving capabilities. Mastering these techniques is essential for success in mathematics and its applications.