Solve Logarithmic Equations: Log 4 + Log(x+2) = 1

Hey math whizzes and algebra adventurers! Today, we're diving deep into the cool world of logarithms to tackle a problem that might look a little intimidating at first glance, but trust me, guys, it's totally conquerable. We're going to figure out which equation is equivalent to the logarithmic equation: $\log 4 + \log (x+2) = 1$. This isn't just about getting the right answer; it's about understanding the why and the how behind solving logarithmic equations. We'll break down the properties of logarithms and see how they transform our original equation into a simpler, more manageable form. So, grab your calculators (or just your brilliant brains!), and let's get this done.

Understanding the Basics of Logarithms

Before we jump into solving, let's quickly refresh our memories on what logarithms are all about. The equation $\log_b a = c$ is essentially asking: "To what power ($c$) must we raise the base ($b$) to get the number ($a$)?" In simpler terms, it's the inverse operation of exponentiation. When you see just "log" without a base specified, it usually means the common logarithm, which has a base of 10 (i.e., $\log_{10}$). So, our equation $\log 4 + \log (x+2) = 1$ is actually $\log_{10} 4 + \log_{10} (x+2) = 1$. Understanding this base is crucial because it dictates how we'll convert our logarithmic equation into an algebraic one. Remember, the properties of logarithms are our best friends here. The product rule states that $\log_b M + \log_b N = \log_b (M \times N)$. This is the key property we'll be using to combine the two logarithmic terms on the left side of our equation. Applying this rule, $\log 4 + \log (x+2)$ becomes $\log (4 \times (x+2))$. So, our equation transforms into $\log (4(x+2)) = 1$. This is a much cleaner form, bringing us one step closer to the solution. It's like finding a shortcut on a long road trip! This step alone simplifies the problem significantly and highlights the power of using logarithmic properties. We're not changing the core problem; we're just rewriting it in a more convenient form that allows us to proceed with solving for $x$. Always keep an eye out for opportunities to use these properties, guys, they're lifesavers!

Applying Logarithm Properties to Simplify

So, we've established that $\log 4 + \log (x+2)$ can be rewritten as $\log (4(x+2))$ using the product rule for logarithms. Our equation now stands as $\log (4(x+2)) = 1$. This simplified form is fantastic because it isolates the logarithmic expression on one side. Now, remember that 'log' without a base implies base 10. So, we have $\log_{10} (4(x+2)) = 1$. To get rid of the logarithm and solve for $x$, we need to convert this logarithmic equation into its equivalent exponential form. The general rule for converting from logarithmic to exponential form is: if $\log_b y = x$, then $b^x = y$. Applying this to our equation, where our base $b=10$, our exponent $x=1$, and our argument $y = 4(x+2)$, we get $10^1 = 4(x+2)$. This is the exact form we are looking for! We have successfully transformed the original logarithmic equation into an algebraic equation that directly relates the base, exponent, and argument. This is the essence of solving logarithmic equations – using their properties to simplify and then converting them to a solvable form, usually an exponential or a linear/quadratic equation. The expression $10^1 = 4(x+2)$ is what we need to match against the given options. We've done the heavy lifting by correctly applying the product rule and the definition of a logarithm. It's all about manipulating the equation using established mathematical rules. This step is vital because it removes the logarithm entirely, allowing us to treat the equation as a standard algebraic problem.

Converting to Exponential Form and Finding the Equivalent Equation

We've arrived at the crucial step where we have $\log (4(x+2)) = 1$. As established, the base of this logarithm is 10. The definition of a logarithm tells us that $\log_b a = c$ is equivalent to $b^c = a$. In our case, the base $b=10$, the exponent $c=1$, and the argument $a = 4(x+2)$. Therefore, converting our equation $\log_{10} (4(x+2)) = 1$ to its exponential form gives us $10^1 = 4(x+2)$. This equation, $4(x+2) = 10^1$, is now in a form that we can easily solve for $x$. We can distribute the 4 to get $4x + 8 = 10^1$. This matches one of our options perfectly! It's important to note that sometimes the options might present the equation in a slightly different but algebraically equivalent form. For instance, $10^1$ is simply 10. So, the equation could also be written as $4x + 8 = 10$. However, the options provided maintain the $10^1$ format, which is exactly what we derived directly from the logarithmic equation. This step solidifies our understanding that logarithmic equations can be directly translated into exponential equations, and vice-versa. It's a fundamental bridge between two powerful mathematical concepts. We've followed the rules of logarithms precisely, ensuring that our transformation is accurate and leads to a valid equivalent equation. The goal was to find an equation that represents the same relationship between the variables and constants as the original logarithmic equation, and $4x + 8 = 10^1$ does just that.

Analyzing the Options

Now that we've meticulously derived the equivalent exponential equation, it's time to compare our result with the given options. Our derived equation is $4(x+2) = 10^1$, which, after distributing the 4, becomes $4x + 8 = 10^1$. Let's scrutinize each option:

• A. $x-2=10^1$: This equation doesn't reflect the combination of the logarithms or the multiplication by 4. It seems to have ignored the '4' and the structure of the argument. Definitely not equivalent.
• B. $4x+8=10^1$: Bingo! This option precisely matches the equation we derived. It correctly incorporates the product of 4 and $(x+2)$ on the left side and the base raised to the power of 1 on the right side. This is our equivalent equation.
• C. $4x+8=1$: This option is close, but it incorrectly equates $4x+8$ to just '1' instead of $10^1$. This would be the result if the original equation was $\log(4x+8) = 0$ (since $10^0 = 1$), or if the right side of the original equation was $\log 1$. This is a common mistake if one forgets the base of the logarithm or mishandles the conversion.
• D. $\frac{4}{x+2}=10^1$: This equation would arise if the original problem involved division of logarithms (which isn't a standard property like addition or subtraction) or perhaps a subtraction of logarithms, $\log 4 - \log (x+2) = 1$, leading to $\log \left(\frac{4}{x+2}\right) = 1$, and then converting to exponential form. But our original problem had addition.

Therefore, after careful step-by-step derivation and comparison, option B is unequivocally the equation equivalent to $\log 4 + \log (x+2) = 1$. It’s super satisfying when the math lines up perfectly, right? This process really shows how important it is to know and apply the properties of logarithms correctly. Each step, from combining terms to converting to exponential form, is vital in reaching the correct equivalent equation.

Solving for x (Just for Fun!)

While the question only asks for the equivalent equation, let's quickly solve for $x$ from our confirmed equivalent equation, $4x + 8 = 10^1$. We know $10^1 = 10$. So, we have $4x + 8 = 10$. Subtracting 8 from both sides gives us $4x = 10 - 8$, which simplifies to $4x = 2$. Dividing both sides by 4, we get $x = \frac{2}{4}$, or $x = \frac{1}{2}$.

Now, it's always a good practice with logarithmic equations to check if our solution is valid by plugging it back into the original equation. The argument of a logarithm must be positive. In our original equation, $\log 4 + \log (x+2) = 1$, the arguments are 4 (which is positive) and $(x+2)$. If $x = \frac{1}{2}$, then $x+2 = \frac{1}{2} + 2 = \frac{1}{2} + \frac{4}{2} = \frac{5}{2}$. Since $\frac{5}{2}$ is positive, our solution $x = \frac{1}{2}$ is valid.

Let's check the original equation: $\log 4 + \log (\frac{1}{2}+2) = \log 4 + \log (\frac{5}{2})$. Using the product rule, this is $\log (4 \times \frac{5}{2}) = \log (\frac{20}{2}) = \log 10$. Since the base is 10, $\log_{10} 10 = 1$. This matches the right side of the original equation. So, our solution $x=\frac{1}{2}$ is correct, and the equivalent equation $4x+8=10^1$ (option B) is the one that leads us there.

Conclusion

We've successfully navigated the path from a logarithmic equation to its equivalent algebraic form. By applying the product rule of logarithms ($\log M + \log N = \log(MN)$) and understanding the definition of a logarithm ($\log_b a = c \iff b^c = a$), we transformed $\log 4 + \log (x+2) = 1$ into $10^1 = 4(x+2)$, which simplifies to $4x+8=10^1$. This step-by-step process confirms that option B is the correct answer. It's a great example of how mastering fundamental mathematical properties can unlock complex problems. Keep practicing, stay curious, and remember that every equation has a story to tell! You guys crushed it!