Solve $3^{2x}=7^{3-x}$ For X: Logarithmic Solution

Understanding Exponential Equations

Exponential equations can seem a bit daunting at first glance, especially when the variable is in the exponent. But don't worry, they're quite manageable with the right tools! Today, we're going to tackle a specific one: $3^{2x}=7^{3-x}$. Our goal is to find the value of 'x' that makes this equation true. The key to unlocking these types of problems often lies in converting them into a more familiar form, and that's where logarithms come into play. Think of logarithms as the inverse operation of exponentiation – they help us bring those pesky exponents down to a level where we can work with them. We'll be using this powerful relationship to isolate 'x' and discover its value. So, let's dive in and demystify solving exponential equations!

Converting to Logarithmic Form

To solve the exponential equation $3^{2x}=7^{3-x}$ for $x$, the first strategic step is to convert it into logarithmic form. This might sound complicated, but it's a fundamental technique that allows us to manipulate the equation more effectively. The core idea is that if we have an equation of the form $a^b = c$, we can rewrite it in logarithmic form as $\log_a(c) = b$. In our specific equation, $3^{2x}=7^{3-x}$, we have terms with different bases (3 and 7) and different exponents involving 'x'. Directly applying the definition of a logarithm to the entire equation isn't straightforward because we have two separate exponential terms. Instead, a more practical approach is to take the logarithm of both sides of the equation. This preserves the equality. We can choose to use either the common logarithm (base 10, denoted as log) or the natural logarithm (base $e$, denoted as ln). Both will lead us to the correct answer, but the problem specifically asks for the answer to be expressed as a fraction with only one logarithm in the numerator, which is achievable with either.

Let's choose to use the natural logarithm (ln) for this demonstration, though the common logarithm (log) would yield a structurally similar result. Applying the natural logarithm to both sides of $3^{2x}=7^{3-x}$, we get:

$\ln(3^{2x}) = \ln(7^{3-x})$

This step is crucial because it allows us to utilize a fundamental property of logarithms: the power rule. The power rule states that $\ln(a^b) = b \ln(a)$. We can apply this rule to both sides of our equation. On the left side, the exponent is $2x$, and on the right side, the exponent is $3-x$. Applying the power rule, we can bring these exponents down as multipliers of the logarithms:

$$(2x) \ln(3) = (3-x) \ln(7)$$

This transformation is incredibly powerful. We've successfully moved the variable 'x' out of the exponent and into a position where we can treat it as a standard algebraic variable. The equation now involves linear terms with 'x' multiplied by logarithmic constants ($\ln(3)$ and $\ln(7)$). The next phase of our problem-solving journey will focus on isolating 'x' from this new form of the equation.

Isolating the Variable 'x'

Now that we have transformed our exponential equation $3^{2x}=7^{3-x}$ into $(2x) \ln(3) = (3-x) \ln(7)$ using logarithmic properties, our primary objective is to isolate the variable 'x'. This stage involves standard algebraic manipulation, similar to solving any linear equation. The 'x' terms are currently on both sides of the equation, each multiplied by a logarithmic constant. Our strategy will be to gather all terms containing 'x' on one side of the equation and all constant terms on the other.

First, let's distribute the logarithms on both sides to clearly see the terms involving 'x':

$$2x \ln(3) = 3 \ln(7) - x \ln(7)$$

Next, we want to move the term $-x \ln(7)$ from the right side to the left side. We can do this by adding $x \ln(7)$ to both sides of the equation:

$$2x \ln(3) + x \ln(7) = 3 \ln(7)$$

Notice how all terms with 'x' are now on the left side. The next crucial step is to factor out 'x' from the terms on the left side. This is a common technique when 'x' appears in multiple additive terms:

$$x(2 \ln(3) + \ln(7)) = 3 \ln(7)$$

We are now very close to isolating 'x'. The variable 'x' is currently being multiplied by the entire expression $(2 \ln(3) + \ln(7))$. To get 'x' by itself, we need to divide both sides of the equation by this expression:

$$x = \frac{3 \ln(7)}{2 \ln(3) + \ln(7)}$$

This equation gives us the exact value of 'x'. However, the problem specifies that the answer should be expressed as a fraction with only one logarithm in the numerator. Our current expression has logarithms in both the numerator and the denominator, and also a coefficient in the denominator. We need to simplify it further using logarithm properties.

Expressing the Answer as a Fraction with One Logarithm

We have arrived at the solution for $x$ as $x = \frac{3 \ln(7)}{2 \ln(3) + \ln(7)}$. The next crucial step, as per the problem's requirements, is to express this answer as a fraction with only one logarithm in the numerator. This means we need to manipulate the denominator using logarithmic properties to simplify it into a single logarithmic term. We can achieve this by leveraging the product rule and the power rule of logarithms.

Recall the power rule for logarithms: $b \ln(a) = \ln(a^b)$. Applying this to the term $2 \ln(3)$ in our denominator, we get:

$$2 \ln(3) = \ln(3^2) = \ln(9)$$

Now, substitute this back into our expression for $x$:

$$x = \frac{3 \ln(7)}{\ln(9) + \ln(7)}$$

Next, we use the product rule for logarithms, which states that $\ln(a) + \ln(b) = \ln(ab)$. Applying this to the denominator $\ln(9) + \ln(7)$, we combine these two logarithms into a single one:

$$\ln(9) + \ln(7) = \ln(9 \times 7) = \ln(63)$$

Substituting this simplified denominator back into our expression for $x$, we get:

$$x = \frac{3 \ln(7)}{\ln(63)}$$

We've successfully isolated 'x' and simplified the expression. However, we still have a coefficient '3' multiplying the logarithm in the numerator. The requirement is for only one logarithm in the numerator. To achieve this, we can again use the power rule of logarithms in reverse: $b \ln(a) = \ln(a^b)$. Applying this to the numerator $3 \ln(7)$, we get:

$$3 \ln(7) = \ln(7^3)$$

Now, we calculate $7^3$: $7 \times 7 = 49$, and $49 \times 7 = 343$. So, $7^3 = 343$. Therefore, the numerator becomes:

$$\ln(343)$$

Finally, substitute this back into our expression for $x$, and we arrive at the desired form:

$$x = \frac{\ln(343)}{\ln(63)}$$

This expression satisfies all the conditions: it provides the exact solution for $x$, it is a fraction, and it has only one logarithm in the numerator (and also only one in the denominator, though the primary constraint was on the numerator). This form is also convenient for calculating a numerical approximation using a calculator.

Conclusion: Mastering Exponential Equations

We've successfully navigated the process of solving the exponential equation $3^{2x}=7^{3-x}$ for $x$, culminating in an answer expressed as a fraction with only one logarithm in the numerator: $x = \frac{\ln(343)}{\ln(63)}$. This journey involved key steps: first, taking the logarithm of both sides of the equation to bring the exponents down, then applying the power rule of logarithms. Subsequently, we used algebraic manipulation to gather terms with 'x' on one side and constants on the other, followed by factoring out 'x'. The final crucial phase involved simplifying the expression using the power rule and product rule of logarithms to meet the specific format requirement of having a single logarithm in the numerator. This problem highlights the power and versatility of logarithms in transforming complex exponential relationships into solvable algebraic forms.

Remember, the ability to convert between exponential and logarithmic forms, coupled with a solid understanding of logarithmic properties (power rule, product rule, quotient rule), are essential tools in your mathematics arsenal. Whether you're dealing with simple or complex exponential equations, these techniques will serve you well. Practice is key; the more you work through different types of problems, the more intuitive these steps will become. You might find it helpful to explore further resources on solving logarithmic and exponential equations to solidify your understanding.

For additional insights and practice on logarithms and exponential functions, I recommend checking out the resources at Khan Academy or the MathWorld website. These sites offer comprehensive explanations, examples, and practice exercises that can further enhance your understanding of these fundamental mathematical concepts. They are excellent starting points for anyone looking to deepen their knowledge in mathematics.