Solve Log Equations Graphically: A Step-by-Step Guide

Hey guys! Let's dive into a cool math problem Tenisha tackled using graphs. She had to solve the equation:

$$\log _3 5 x=\log _5(2 x+8)$$

To solve this, Tenisha cleverly graphed a system of equations. Our mission? To pinpoint the point that approximates the solution for her system. We've got these options to choose from:

(0. 9, 0.8) (1. 0, 1.4) (2. 3, 1.1) (2. 7, 13.3)

Let's break it down step by step to figure out the correct answer!

Understanding the Problem: Logarithmic Equations and Graphical Solutions

Before we jump into the solution, let's make sure we're all on the same page. We're dealing with a logarithmic equation here. Remember, logarithms are just the inverse of exponentiation. The equation Tenisha faced has logarithms with different bases (base 3 and base 5), which makes solving it algebraically a bit tricky. That's why she opted for a graphical approach – a smart move!

The key idea behind solving equations graphically is to transform the single equation into a system of two equations. We do this by setting each side of the original equation equal to 'y'. So, in Tenisha's case, she likely created the following system:

• y = log₃(5x)
• y = log₅(2x + 8)

The solution to the original equation is the x-coordinate of the point where the graphs of these two equations intersect. Why? Because at the point of intersection, the y-values are equal, meaning both sides of the original equation are equal.

Now, let's consider what we're looking for. We need to find a point (x, y) that approximately lies on both curves. This means when we plug the x-value into both logarithmic functions, we should get approximately the same y-value. We will need to estimate these values since we do not have the luxury of graphing the equations.

Let's get our hands dirty and test the given points!

Testing the Options: Finding the Intersection Point

Okay, team, let's put on our detective hats and examine each point to see which one fits the bill. We'll plug the x-value of each point into both equations (y = log₃(5x) and y = log₅(2x + 8)) and see if the resulting y-values are close.

Option 1: (0.9, 0.8)

• For y = log₃(5x): y ≈ log₃(5 * 0.9) = log₃(4.5). To estimate this, we know 3¹ = 3 and 3². = 9, so log₃(4.5) will be somewhere between 1 and 2, closer to 1 (since 4.5 is closer to 3 than 9). A rough estimate would be around 1.4. This is significantly different from the y-value of the point (0.8).
• For y = log₅(2x + 8): y ≈ log₅(2 * 0.9 + 8) = log₅(9.8). Since 5¹ = 5 and 5². = 25, log₅(9.8) is between 1 and 2, closer to 1. A rough estimate would be around 1.4.

The y-values don't seem to match closely, and one of the calculated y-values doesn't even match the y-coordinate of the point. So, (0.9, 0.8) is likely not the solution.

Option 2: (1.0, 1.4)

• For y = log₃(5x): y ≈ log₃(5 * 1.0) = log₃(5). Again, 3¹ = 3 and 3² = 9, so log₃(5) is between 1 and 2. It's closer to 1 than 2. A rough estimate might be around 1.5. This is reasonably close to the y-value of 1.4 in the point.
• For y = log₅(2x + 8): y ≈ log₅(2 * 1.0 + 8) = log₅(10). Since 5¹ = 5 and 5² = 25, log₅(10) is between 1 and 2, leaning more towards 1. We can estimate this to be around 1.4.

Wow! The calculated y-values (1.5 and 1.4) are pretty close to each other, and close to the y-coordinate of the point (1.4). This looks promising!

Option 3: (2.3, 1.1)

• For y = log₃(5x): y ≈ log₃(5 * 2.3) = log₃(11.5). Since 3² = 9 and 3³ = 27, log₃(11.5) is between 2 and 3, closer to 2. We might estimate this around 2.2. This is quite different from 1.1, the y-value of the point.
• For y = log₅(2x + 8): y ≈ log₅(2 * 2.3 + 8) = log₅(12.6). Since 5¹ = 5 and 5² = 25, log₅(12.6) is between 1 and 2. We estimate around 1.5.

The calculated y-values are quite different from each other and from the point's y-coordinate. So, (2.3, 1.1) is not our solution.

Option 4: (2.7, 13.3)

• For y = log₃(5x): y ≈ log₃(5 * 2.7) = log₃(13.5). This is between 2 and 3 (since 3² = 9 and 3³ = 27). A rough estimate is around 2.4.
• For y = log₅(2x + 8): y ≈ log₅(2 * 2.7 + 8) = log₅(13.4). This is also between 1 and 2 (since 5¹ = 5 and 5² = 25). We can estimate this as around 1.6.

The calculated y-values are not close to each other, and neither is close to 13.3. This option is definitely not the solution.

The Verdict: Our Approximate Solution

Drumroll, please! After carefully analyzing each option, it's clear that the point (1.0, 1.4) is the best approximation for the solution to Tenisha's system of equations. The y-values we calculated when plugging x = 1.0 into both equations were very close to each other and reasonably close to the y-coordinate of the point itself.

Key Takeaways: Graphical Solutions and Logarithms

Great job, team! We successfully found the approximate solution by understanding the power of graphical methods. Let's recap the key concepts we used:

• Graphical Solutions: Transforming a single equation into a system of two equations (by setting each side equal to 'y') allows us to find solutions by identifying intersection points.
• Logarithms: Remembering the relationship between logarithms and exponents helps us estimate logarithmic values without a calculator.
• Approximation: When dealing with graphical solutions, we're often looking for approximate solutions, as reading exact values from a graph can be challenging.

This problem highlights a powerful problem-solving strategy in mathematics: when faced with a tricky equation, consider visual representations like graphs. They can often provide valuable insights and lead you to the solution!