Unlocking Solutions: A Guide To Solving Logarithmic Equations

Hey math enthusiasts! Ready to dive into the world of logarithms and conquer some equations? Today, we're going to break down how to solve a couple of logarithmic equations. Don't worry, it's not as scary as it sounds! We'll go step-by-step, making sure you grasp the concepts and techniques. We'll look at the equations: a) $\ln (2 x+1)-\ln 4=\ln x$ and b) $\log _3 x+\log _3(x+8)=2$. Let's get started!

Solving Equation a) $\ln (2 x+1)-\ln 4=\ln x$

Alright, guys, let's tackle the first equation. We've got $\ln (2 x+1)-\ln 4=\ln x$. Our goal is to isolate x. The key here is to use the properties of logarithms. Remember, the difference of two logarithms is the same as the logarithm of the quotient. So, we can rewrite the left side of the equation. First, we will be going through the properties of logarithms to unlock the solution. This will help you easily solve any kind of equation!

Using the quotient rule of logarithms, which states that $\ln a - \ln b = \ln(a/b)$, we can combine the two logarithmic terms on the left side: $\ln((2x+1)/4) = \ln x$. Now, since we have the natural logarithm (ln) on both sides, we can equate the arguments (the expressions inside the logarithms). This is a crucial step! So, we get $(2x+1)/4 = x$. Now we have eliminated the logarithms, which makes our life easier. It is now just a simple algebraic equation. To solve for x, multiply both sides by 4 to get rid of the fraction: $2x + 1 = 4x$. Subtract $2x$ from both sides: $1 = 2x$. Finally, divide by 2: $x = 1/2$. But wait, we're not done yet! We always need to check our answer to make sure it's valid. This is an important step to make sure you have the right answer. We need to plug our solution back into the original equation to ensure it doesn't cause any issues, such as taking the logarithm of a negative number or zero. Let's plug $x = 1/2$ back into the original equation: $\ln (2*(1/2) + 1) - \ln 4 = \ln (1/2)$. This simplifies to $\ln 2 - \ln 4 = \ln (1/2)$. Remember, $\ln 2 - \ln 4$ can be rewritten as $\ln(2/4)$, which is $\ln(1/2)$. So, $\ln(1/2) = \ln(1/2)$. The solution checks out! Thus, the solution to the first equation is $x = 1/2$. Easy, right? We have used various properties of logarithms like quotient rules to solve this question. The most important thing in this step is to understand the properties of logarithms. This helps you to solve any kind of question easily. This method helps you to break down complex equations.

The Importance of Checking Your Solution

Why do we always check our answer? Well, it's because logarithms have restrictions. You can't take the logarithm of a negative number or zero. When we solve logarithmic equations, we might sometimes end up with solutions that don't make sense in the original equation. That is why it is super important to check our solution every time. This is why checking our solution is always super important. In our case, $x=1/2$ is perfectly fine because we don't end up taking the logarithm of a negative number or zero. So, always remember to verify your solutions!

Solving Equation b) $\log _3 x+\log _3(x+8)=2$

Now, let's move on to the second equation: $\log _3 x+\log _3(x+8)=2$. This time, we have logarithms with a base of 3. We'll use a similar approach. The key here is the product rule of logarithms. When we add the logarithms, we can combine the arguments by multiplying them. So, we can combine the two logarithmic terms on the left side using the product rule, which states that $\log_b a + \log_b c = \log_b(a*c)$. This gives us $\log_3(x * (x+8)) = 2$. Now, we need to get rid of the logarithm. To do this, we can rewrite the equation in exponential form. Remember that $\log_b a = c$ is equivalent to $b^c = a$. Applying this to our equation, we get $3^2 = x(x+8)$. This simplifies to $9 = x^2 + 8x$. Now we have a quadratic equation! To solve for x, we need to rearrange the equation into standard quadratic form, which is $ax^2 + bx + c = 0$. Subtract 9 from both sides: $x^2 + 8x - 9 = 0$. Next, we need to factor the quadratic equation. We're looking for two numbers that multiply to -9 and add up to 8. Those numbers are 9 and -1. So, we can factor the equation as $(x+9)(x-1) = 0$. Now we can set each factor equal to zero and solve for x: $x + 9 = 0$ gives us $x = -9$, and $x - 1 = 0$ gives us $x = 1$. Again, before we declare victory, we need to check our solutions to make sure they're valid. Remember, we cannot take the logarithm of a negative number or zero. Let's plug each solution back into the original equation.

Validating the Solution

First, let's check $x = -9$. If we plug this into the original equation, we get $\log_3(-9) + \log_3(-9+8) = 2$, which simplifies to $\log_3(-9) + \log_3(-1) = 2$. Uh oh! We can't take the logarithm of a negative number. So, $x = -9$ is not a valid solution. It is called an extraneous solution. Now, let's check $x = 1$. Plugging this into the original equation, we get $\log_3 1 + \log_3(1+8) = 2$, which simplifies to $\log_3 1 + \log_3 9 = 2$. We know that $\log_3 1 = 0$ and $\log_3 9 = 2$, so we have $0 + 2 = 2$. This is true! So, the only valid solution to the second equation is $x = 1$. This is the final answer! Always remember to check your answers when dealing with logarithms! It helps you to avoid common mistakes, and the solutions must be validated. It also helps to eliminate any invalid solutions.

Conclusion: Mastering Logarithmic Equations

Alright, guys, we've successfully navigated through two logarithmic equations! We've seen how to use the properties of logarithms, such as the quotient rule and the product rule, to simplify equations. We've also learned the importance of checking our solutions to ensure they're valid. Remember that understanding the properties of logarithms and always checking your solutions is the key to mastering these types of problems. Keep practicing, and you'll become a pro in no time! Keep in mind that math is all about practice. So do a lot of practice to get better at solving these questions. The more questions you solve, the easier it becomes! That's all for now. Feel free to reach out if you have any questions or want to explore more examples. Happy solving!