Graphing Logarithmic Equations: Find The Intersection Point

Hey there, math enthusiasts! Today, we're diving into the fascinating world of logarithmic equations and exploring how we can solve them not just with algebra, but with the visual power of graphing. Our specific challenge is to determine the truth behind the intersection point of the curves represented by the equation $\log (6 x+10)=\log _{\frac{1}{2}} x$. This isn't just about finding a number; it's about understanding the relationship between different logarithmic functions and where they meet. We'll be looking at statements suggesting intersections around $x=0.46$, $x=0.75$, and $x=1.21$. Let's get our graphing tools ready and uncover which statement holds true!

Understanding the Equation and Graphing Concepts

Before we jump into graphing, let's break down the equation $\log (6 x+10)=\log _{\frac{1}{2}} x$. We have two logarithmic expressions set equal to each other. The first, $\log (6 x+10)$, is a common logarithm (base 10, usually written without a subscript). For this term to be defined, the argument must be positive, meaning $6x + 10 > 0$, which simplifies to $x > -10/6$ or $x > -5/3$. The second term, $\log _{\frac{1}{2}} x$, is a logarithm with a fractional base ($1/2$). For this term to be defined, the argument $x$ must also be positive, so $x > 0$. Combining these conditions, the domain for our solution is $x > 0$. This is a crucial first step, as any potential solutions outside this range are invalid.

When we solve an equation by graphing, we're essentially visualizing each side of the equation as a separate function. So, we'll consider $y_1 = \log (6 x+10)$ and $y_2 = \log _{\frac{1}{2}} x$. The solutions to the original equation are the $x$-values where the graphs of $y_1$ and $y_2$ intersect. Graphing these functions helps us to intuitively see where the equality holds. The common logarithm, $y_1 = \log (6 x+10)$, will generally increase as $x$ increases. The logarithmic function with a base between 0 and 1, $y_2 = \log _{\frac{1}{2}} x$, is a decreasing function. This difference in behavior is important. As $x$ increases, $y_1$ increases, and $y_2$ decreases. This means that if they intersect, they are likely to do so at only one point. The question gives us specific $x$-values where we are told the curves might intersect: approximately $0.46$, $0.75$, and $1.21$. Our task is to graph these functions and see which of these $x$-values corresponds to an intersection point.

To graph these accurately, we can use a graphing calculator or online graphing tools. We need to input the functions $y = \log(6x+10)$ and $y = \log_{1/2}(x)$. Remember that graphing calculators often don't have a direct button for $\log_{1/2}(x)$. In such cases, we can use the change of base formula for logarithms: $\log_b a = \frac{\log a}{\log b}$ (using common logs) or $\log_b a = \frac{\ln a}{\ln b}$ (using natural logs). So, $\log_{\frac{1}{2}} x$ can be rewritten as $\frac{\log x}{\log (1/2)}$ or $\frac{\ln x}{\ln (1/2)}$. Since $\log (1/2) = \log (2^{-1}) = -\log 2$ and $\ln (1/2) = \ln (2^{-1}) = -\ln 2$, we can write $\log_{\frac{1}{2}} x = \frac{\log x}{-\log 2}$ or $\frac{\ln x}{-\ln 2}$. This rewritten form is essential for accurate plotting. As we plot, we must keep in mind our domain restriction $x > 0$. The graphs will only be meaningful for positive $x$-values.

Analyzing the Intersection Points

Let's consider the potential intersection points given: $x \approx 0.46$, $x \approx 0.75$, and $x \approx 1.21$. To verify which of these is true, we can substitute these values back into the original equation or, more visually, check if the $y$-values are approximately equal when graphed. For instance, if we pick $x = 0.46$, we would calculate $y_1 = \log (6 \times 0.46 + 10)$ and $y_2 = \log _{\frac{1}{2}} 0.46$. If $y_1 \approx y_2$, then $x = 0.46$ is a valid intersection point. We need to do this for all given options.

Let's start with $x = 0.46$:

For $y_1 = \log (6x + 10)$: $y_1 = \log (6 \times 0.46 + 10) = \log (2.76 + 10) = \log (12.76)$. Using a calculator, $\log (12.76) \approx 1.1058$.

For $y_2 = \log _{\frac{1}{2}} x$: $y_2 = \log _{\frac{1}{2}} 0.46$. Using the change of base formula: $y_2 = \frac{\log 0.46}{\log (0.5)} \approx \frac{-0.3372}{-0.3010} \approx 1.1203$.

Comparing $y_1 \approx 1.1058$ and $y_2 \approx 1.1203$, these values are close but not exactly the same. This suggests that $x = 0.46$ might be an approximation, but let's check the other values to be sure. Often, when solving by graphing, we're looking for approximate solutions, and the wording of the options implies this.

Now, let's test $x = 0.75$:

For $y_1 = \log (6x + 10)$: $y_1 = \log (6 \times 0.75 + 10) = \log (4.5 + 10) = \log (14.5)$. Using a calculator, $\log (14.5) \approx 1.1614$.

For $y_2 = \log _{\frac{1}{2}} x$: $y_2 = \log _{\frac{1}{2}} 0.75$. Using the change of base formula: $y_2 = \frac{\log 0.75}{\log (0.5)} \approx \frac{-0.1249}{-0.3010} \approx 0.4149$.

Here, $y_1 \approx 1.1614$ and $y_2 \approx 0.4149$. These values are significantly different, so $x = 0.75$ is not the intersection point.

Finally, let's test $x = 1.21$:

For $y_1 = \log (6x + 10)$: $y_1 = \log (6 \times 1.21 + 10) = \log (7.26 + 10) = \log (17.26)$. Using a calculator, $\log (17.26) \approx 1.2370$.

For $y_2 = \log _{\frac{1}{2}} x$: $y_2 = \log _{\frac{1}{2}} 1.21$. Using the change of base formula: $y_2 = \frac{\log 1.21}{\log (0.5)} \approx \frac{0.0828}{-0.3010} \approx -0.2751$.

Here, $y_1 \approx 1.2370$ and $y_2 \approx -0.2751$. These values are also very different, ruling out $x = 1.21$ as the intersection point.

Revisiting our calculation for $x \approx 0.46$, we found $y_1 \approx 1.1058$ and $y_2 \approx 1.1203$. The difference is about $0.0145$. Let's re-evaluate the accuracy. When solving by graphing, especially with approximations, we often look for the closest match. The nature of logarithmic functions means small changes in $x$ can lead to noticeable changes in $y$. However, the initial calculation for $x=0.46$ showed the closest values among the options provided.

To be absolutely certain, let's consider the exact solution algebraically. While the question asks for a graphical solution, understanding the algebraic solution can confirm our graphical findings. We have $\log (6x+10) = \log_{\frac{1}{2}} x$. Let's use the change of base formula for the right side: $\log (6x+10) = \frac{\log x}{\log (1/2)}$. Let $b$ be the base of the common logarithm (usually 10). Then $\log (6x+10) = \frac{\log x}{-\log 2}$. This can be rewritten as $-\log 2 \times \log (6x+10) = \log x$. Using logarithm properties, $\log ((6x+10)^{-\log 2}) = \log x$. This implies $(6x+10)^{-\log 2} = x$. This is a transcendental equation and is very difficult to solve analytically. This is precisely why graphical methods are so valuable for such problems.

Given the difficulty in analytical solution and the nature of the question presenting approximate values, we must rely on accurate graphing or numerical evaluation of the options. Let's refine the calculation for $x = 0.46$ to a higher precision.

Let $x = 0.461$: $y_1 = \log(6 imes 0.461 + 10) = \log(2.766 + 10) = \log(12.766) \approx 1.10606$ $y_2 = \log_{1/2}(0.461) = \frac{\log(0.461)}{\log(0.5)} \approx \frac{-0.33626}{-0.30103} \approx 1.11703$

The difference is still noticeable. Let's try a value closer to $0.46$, perhaps slightly larger, considering that $y_2$ is slightly larger than $y_1$ at $x=0.46$. Let's test $x=0.45$: $y_1 = \log(6 imes 0.45 + 10) = \log(2.7 + 10) = \log(12.7) \approx 1.1038$ $y_2 = \log_{1/2}(0.45) = \frac{\log(0.45)}{\log(0.5)} \approx \frac{-0.34679}{-0.30103} \approx 1.1520$

The difference is larger. This suggests that the intersection point is indeed close to $0.46$, and our initial evaluation was likely the most accurate among the choices. The wording "approximately" in the options is key here.

Let's look at the graphical representation. The function $y = \log(6x+10)$ starts at $x$ slightly greater than $-5/3$ and increases. The function $y = \log_{1/2}(x)$ starts at $x>0$ and decreases. At $x$ very close to 0 (but positive), $\log(6x+10)$ will be close to $\log(10) = 1$. However, $\log_{1/2}(x)$ will be very large positive. As $x$ increases, $\log(6x+10)$ increases, and $\log_{1/2}(x)$ decreases. We need to find where these values cross. Let's examine the behavior near $x=0.46$. We saw $y_1 approx 1.1058$ and $y_2 approx 1.1203$. The values are very close. The graphical intersection point is the $x$-value where these two $y$-values are equal. Given the options, $x \approx 0.46$ appears to be the most plausible intersection.

To confirm this, let's consider if there's a point where $y_1$ and $y_2$ are exactly equal. If we set $y_1 = y_2 = k$, then $6x+10 = 10^k$ and $x = (1/2)^k$. Substituting the second into the first: $6(1/2)^k + 10 = 10^k$. This equation is also difficult to solve directly. However, if we consider the values we calculated: at $x=0.46$, $y_1 approx 1.106$ and $y_2 approx 1.120$. The difference is small. If we were to plot these precisely, the intersection would occur where the values match. Based on the provided options and the numerical evaluation, option A seems to be the correct statement.

Visualizing the Intersection with a Graphing Tool

To truly solidify our understanding, let's imagine using a graphing utility. When we plot $y = \log(6x+10)$ and $y = \log_{1/2}(x)$ on the same axes, we'll observe their behavior. The first graph, $y_1$, will rise from the left (for $x>-5/3$) and continue upwards. The second graph, $y_2$, will start from very high positive $y$-values as $x$ approaches 0 from the right, and then decrease steadily as $x$ increases. Somewhere in the first quadrant (since $x>0$), these two curves must cross. We are looking for the $x$-coordinate of this crossing point.

Using a tool like Desmos or a TI-84 calculator, inputting y = log(6x+10) and y = log(x)/log(0.5) will generate the visual representation. If you zoom in around $x=0.46$, you'll see the two curves getting very close to each other. By using the calculator's