Solving Logarithmic Equations: Find Exact Value Of X

Let's dive into solving a logarithmic equation step-by-step. Our goal is to find the exact value of x in the equation: $\log _6(3 x)+2 \log _6(3)=4$. Logarithmic equations might seem daunting at first, but with a solid understanding of logarithmic properties and algebraic manipulation, they become quite manageable. We'll break down each step to ensure clarity and comprehension. Let's begin by understanding the fundamental properties of logarithms that we'll be using throughout this solution. Remember, logarithms are essentially the inverse of exponentiation. The expression $\log_b(a) = c$ means that $b^c = a$. We'll also utilize the power rule of logarithms, which states that $\log_b(m^n) = n \log_b(m)$, and the product rule, which tells us that $\log_b(mn) = \log_b(m) + \log_b(n)$. Finally, keep in mind that our ultimate goal is to isolate x on one side of the equation. This often involves combining logarithmic terms, converting the equation to exponential form, and performing algebraic simplifications.

Step-by-Step Solution

1. Apply the Power Rule of Logarithms

Our equation is $\log _6(3 x)+2 \log _6(3)=4$. The first thing we'll address is the term $2 \log _6(3)$. Using the power rule of logarithms, which states that $n \log_b(m) = \log_b(m^n)$, we can rewrite this term as $\log _6(3^2)$. Therefore, $2 \log _6(3) = \log _6(9)$. Now our equation looks like this: $\log _6(3 x)+\log _6(9)=4$.

2. Apply the Product Rule of Logarithms

Now we have two logarithmic terms with the same base being added together: $\log _6(3 x)+\log _6(9)=4$. We can use the product rule of logarithms, which states that $\log_b(m) + \log_b(n) = \log_b(mn)$, to combine these two terms into a single logarithm. So, we have $\log _6(3x \cdot 9) = 4$, which simplifies to $\log _6(27x) = 4$.

3. Convert to Exponential Form

To get rid of the logarithm, we need to convert the equation from logarithmic form to exponential form. Recall that $\log_b(a) = c$ is equivalent to $b^c = a$. In our case, we have $\log _6(27x) = 4$, so converting to exponential form gives us $6^4 = 27x$.

4. Simplify and Solve for x

Now we have a simple algebraic equation: $6^4 = 27x$. First, let's calculate $6^4$. $6^4 = 6 \cdot 6 \cdot 6 \cdot 6 = 36 \cdot 36 = 1296$. So, our equation is now $1296 = 27x$. To isolate x, we need to divide both sides of the equation by 27: $x = \frac{1296}{27}$. Now, let's simplify the fraction. We can divide both the numerator and the denominator by 27. $1296 \div 27 = 48$. Therefore, $x = 48$.

Verification

To ensure our solution is correct, we should plug x = 48 back into the original equation: $\log _6(3 x)+2 \log _6(3)=4$. Substituting x = 48, we get $\log _6(3 \cdot 48)+2 \log _6(3)=4$, which simplifies to $\log _6(144)+2 \log _6(3)=4$. We know that $144 = 6^2$, so $\log _6(144) = 2$. Also, $2 \log _6(3) = \log _6(3^2) = \log _6(9)$. So, the equation becomes $2 + \log _6(9) = 4$. Subtracting 2 from both sides gives $\log _6(9) = 2$. To check if this is true, we convert it to exponential form: $6^2=9$, which simplifies to $\log_6(9)$. Let's convert everything to a base of 3. We can rewrite the terms as follows: $\log_6(144) = \frac{\log_3(144)}{\log_3(6)}$ and $2\log_6(3) = 2\frac{\log_3(3)}{\log_3(6)} = \frac{2}{\log_3(6)}$. Notice that $144 = 3^2 * 2^4$, and $6 = 3*2$, we can further break down $\frac{\log_3(144)}{\log_3(6)}$ to $\frac{\log_3(3^2 * 2^4)}{\log_3(3*2)} = \frac{\log_3(3^2) + \log_3(2^4)}{\log_3(3) + \log_3(2)} = \frac{2 + 4\log_3(2)}{1 + \log_3(2)}$. Adding back to the original equation we have $\frac{2 + 4\log_3(2)}{1 + \log_3(2)} + \frac{2}{1 + \log_3(2)} = \frac{4 + 4\log_3(2)}{1 + \log_3(2)} = 4$. Thus we have validated that x = 48.

Conclusion

Therefore, the exact value of x that satisfies the equation $\log _6(3 x)+2 \log _6(3)=4$ is 48. By applying the power and product rules of logarithms, converting the equation to exponential form, and performing basic algebraic manipulations, we successfully isolated x and found its value. Always remember to verify your solution by plugging it back into the original equation to ensure accuracy. Mastering logarithmic equations involves practice and a solid understanding of logarithmic properties. Keep practicing, and you'll become more comfortable with these types of problems! For further learning and exploration, you can visit Khan Academy's section on logarithms.