Solve $\ln 10+\ln \left(x^2+4\right)=\ln 80$

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the fascinating world of logarithms to tackle a problem that might look a little intimidating at first glance, but trust me, it's totally manageable once we break it down. We're going to solve the equation $\ln 10+\ln \left(x^2+4\right)=\ln 80$. This is a classic logarithmic equation that tests your understanding of log properties, and getting a handle on these is super important, not just for math class, but for a bunch of cool applications in science and engineering too. So, let's get our thinking caps on and unravel this equation step-by-step. We'll use the magic of logarithm rules to simplify things and get to the bottom of what $x$ is.

To kick things off, let's talk about the properties of logarithms that are going to be our best friends for this problem. The most crucial one we'll use here is the product rule of logarithms, which states that $\ln a + \ln b = \ln (a \times b)$. See? It's like giving us permission to combine two separate logarithms into one, which is a huge simplification. We'll also be leaning on the fact that if $\ln a = \ln b$, then $a = b$. This is our way of getting rid of the logarithms once we've managed to isolate them on either side of the equation. It’s a bit like saying, if two things are equal to the same natural logarithm, then the things inside those logarithms must be equal to each other. Before we jump into solving, it's also super important to remember that the argument of a logarithm (the stuff inside the parentheses) must be positive. So, for $\ln \left(x^2+4\right)$, we need $x^2+4 > 0$. Since $x^2$ is always greater than or equal to zero for any real number $x$, $x^2+4$ will always be greater than zero. This means we don't have to worry about any weird domain restrictions popping up later on; our solutions will be valid as long as they are real numbers.

Alright, let's get down to business with our equation: $\ln 10+\ln \left(x^2+4\right)=\ln 80$. Our first move is to use that sweet product rule we just talked about. On the left side of the equation, we have the sum of two natural logarithms. Applying the rule, we can combine them into a single logarithm: $\ln \left(10 \times \left(x^2+4\right)\right)=\ln 80$. Now, let's simplify the expression inside the logarithm on the left. Distributing the 10, we get $\ln \left(10x^2+40\right)=\ln 80$. Look at that! We've successfully combined the left side into a single logarithm. This is exactly where we wanted to be. The equation now looks much cleaner and more approachable. We have a single logarithm on the left and a single logarithm on the right. This sets us up perfectly for the next step, where we'll use the property that allows us to equate the arguments of equal logarithms. Keep your eyes peeled, because the next part is where we start to isolate $x$ and find our actual numerical answers!

So, we're at the stage where we have $\ln \left(10x^2+40\right)=\ln 80$. Remember that golden rule? If $\ln a = \ln b$, then $a = b$. This is our golden ticket to ditching the logarithms entirely and working with a simple algebraic equation. Applying this rule, we can simply set the expressions inside the logarithms equal to each other: $10x^2+40 = 80$. See how much simpler that is? Now, this is a standard quadratic equation, and we know just how to solve these bad boys. Our goal is to isolate $x^2$ first. So, we'll subtract 40 from both sides of the equation: $10x^2 = 80 - 40$, which simplifies to $10x^2 = 40$. The next step to get $x^2$ by itself is to divide both sides by 10: $x^2 = \frac{40}{10}$. Performing that division, we get $x^2 = 4$. Now we're really close! The equation $x^2 = 4$ means we are looking for a number (or numbers) that, when multiplied by itself, equals 4. This is where we take the square root of both sides. Remember, when you take the square root of a number in an equation like this, you have to consider both the positive and negative roots. So, $x = \pm \sqrt{4}$. And what is the square root of 4, you ask? It's 2!

Therefore, our solutions for $x$ are $x = 2$ and $x = -2$. These are the two values that satisfy the original logarithmic equation. To be absolutely sure, especially in math, it's always a smart move to plug these values back into the original equation and check if they work. Let's check $x=2$: $\ln 10+\ln \left(2^2+4\right) = \ln 10+\ln \left(4+4\right) = \ln 10+\ln 8$. Using the product rule again, this becomes $\ln \left(10 \times 8\right) = \ln 80$. Boom! It matches the right side of the original equation. Now let's check $x=-2$: $\ln 10+\ln \left((-2)^2+4\right) = \ln 10+\ln \left(4+4\right) = \ln 10+\ln 8$. This also simplifies to $\ln \left(10 \times 8\right) = \ln 80$. Perfect! Both $x=2$ and $x=-2$ are valid solutions. So, there you have it, guys! We've successfully solved the logarithmic equation $\ln 10+\ln \left(x^2+4\right)=\ln 80$ using the properties of logarithms and a bit of algebraic manipulation. The solutions are $x=2$ and $x=-2$. Remember, practice makes perfect, so keep working through these kinds of problems, and you'll become a log master in no time. Until next time, stay curious and keep exploring the amazing world of math!