Solving $3^{2x+1} = 2^{2x+1}$ An Exponential Equation

In the realm of mathematics, exponential equations often present intriguing challenges. These equations, characterized by variables in their exponents, require a strategic approach to unravel their solutions. This article delves into a specific exponential equation: $3^{2x+1} = 2^{2x+1}$. We will explore the underlying principles and apply various techniques to arrive at a comprehensive solution, elucidating each step along the way.

Understanding Exponential Equations

Exponential equations are mathematical statements where the variable appears in the exponent. The fundamental form of an exponential equation is $a^x = b$, where 'a' is the base, 'x' is the exponent (which contains the variable), and 'b' is the result. Solving these equations involves finding the value(s) of 'x' that satisfy the equation. A key property that governs exponential equations is that if $a^m = a^n$, then $m = n$, provided that 'a' is a positive number not equal to 1. This property allows us to equate the exponents when the bases are the same. However, in the given equation, the bases are different (3 and 2), requiring a different strategy. One approach is to manipulate the equation to have a common base or to utilize logarithms. Logarithms, being the inverse operation of exponentiation, are instrumental in solving exponential equations. The logarithm of a number to a given base is the exponent to which the base must be raised to produce that number. For instance, if $a^x = b$, then $log_a(b) = x$. Logarithms come in different forms, including the common logarithm (base 10) and the natural logarithm (base 'e').

Analyzing the Equation: $3^{2x+1} = 2^{2x+1}$

The exponential equation at hand is $3^{2x+1} = 2^{2x+1}$. A direct application of the property of equating exponents is not feasible here since the bases, 3 and 2, are distinct. Instead, we employ a strategy to manipulate the equation into a more manageable form. One such strategy is to divide both sides of the equation by a common factor. In this case, we can divide both sides by $2^{2x+1}$. This maneuver is valid as $2^{2x+1}$ is always positive for any real value of 'x', ensuring that we are not dividing by zero. The resulting equation is:

$$\frac{3^{2x+1}}{2^{2x+1}} = 1$$

This step is crucial as it allows us to combine the terms with the variable 'x' in the exponent. Recognizing that the exponents are the same, we can apply the rule $\frac{a^n}{b^n} = (\frac{a}{b})^n$. Applying this rule, the equation transforms into:

$$(\frac{3}{2})^{2x+1} = 1$$

Now, the equation has a single base raised to a power involving 'x', making it easier to analyze. The next step involves leveraging the property that any non-zero number raised to the power of zero is equal to 1. This property is key to solving the equation.

Solving for x

Now, we have the equation $(\frac{3}{2})^{2x+1} = 1$. The crucial observation here is that any non-zero number raised to the power of 0 equals 1. Therefore, for the equation to hold true, the exponent $(2x+1)$ must be equal to 0. This leads us to the equation:

$$2x + 1 = 0$$

This is a simple linear equation in 'x', which can be easily solved. To isolate 'x', we first subtract 1 from both sides of the equation:

$$2x = -1$$

Next, we divide both sides by 2 to obtain the value of 'x':

$$x = -\frac{1}{2}$$

Thus, the solution to the exponential equation $3^{2x+1} = 2^{2x+1}$ is $x = -\frac{1}{2}$. To verify this solution, we can substitute $x = -\frac{1}{2}$ back into the original equation:

$$3^{2(-\frac{1}{2})+1} = 3^{-1+1} = 3^0 = 1$$

$$2^{2(-\frac{1}{2})+1} = 2^{-1+1} = 2^0 = 1$$

Since both sides of the equation equal 1 when $x = -\frac{1}{2}$, the solution is verified.

Alternative Approaches: Logarithms

While we solved the exponential equation without explicitly using logarithms, it's worth noting that logarithms offer an alternative approach. We could have taken the logarithm of both sides of the original equation. For instance, taking the natural logarithm (ln) of both sides of $3^{2x+1} = 2^{2x+1}$ gives:

$$ln(3^{2x+1}) = ln(2^{2x+1})$$

Using the power rule of logarithms, which states that $ln(a^b) = b \cdot ln(a)$, we can rewrite the equation as:

$$(2x+1)ln(3) = (2x+1)ln(2)$$

Subtracting $(2x+1)ln(2)$ from both sides gives:

$$(2x+1)ln(3) - (2x+1)ln(2) = 0$$

Factoring out $(2x+1)$:

$$(2x+1)(ln(3) - ln(2)) = 0$$

This equation is satisfied if either $(2x+1) = 0$ or $(ln(3) - ln(2)) = 0$. We already know that $2x+1=0$ leads to $x = -\frac{1}{2}$. The term $(ln(3) - ln(2))$ is non-zero, as $ln(3) \neq ln(2)$. Thus, the only solution is $x = -\frac{1}{2}$, consistent with our previous solution. This logarithmic approach further solidifies the solution and demonstrates the versatility of logarithmic properties in solving exponential equations.

Conclusion

In summary, the exponential equation $3^{2x+1} = 2^{2x+1}$ can be effectively solved by manipulating the equation to equate exponents. By dividing both sides by $2^{2x+1}$, we transformed the equation into $(\frac{3}{2})^{2x+1} = 1$. Recognizing that any non-zero number raised to the power of zero equals 1, we deduced that $2x+1 = 0$, leading to the solution $x = -\frac{1}{2}$. This solution was further validated by substituting it back into the original equation and by employing a logarithmic approach. Exponential equations are a fundamental concept in mathematics, and mastering the techniques to solve them is crucial for various applications in science and engineering. The strategic use of algebraic manipulation and logarithmic properties allows us to tackle these equations with confidence and precision. This exploration highlights the interconnectedness of mathematical concepts and the power of applying fundamental principles to solve complex problems. The solution $x = -\frac{1}{2}$ represents the point where the exponential functions $3^{2x+1}$ and $2^{2x+1}$ intersect, showcasing a key aspect of graphical interpretation in mathematics.