Problem 60 - College Algebra

Question

For the following exercises, solve the equation for \(x\) , if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. $$ \ln (2 x+9)=\ln (-5 x) $$

Step-by-Step Solution

Answer

The solution is \(x = -\frac{9}{7}\). This satisfies the domain conditions, and both log functions intersect at this point.

Step 1: Set the Arguments Equal

Since the logarithm functions are the same (, both with base e), we can set the arguments equal to each other: \[ 2x + 9 = -5x \]

Step 2: Rearrange Terms

Add \(5x\) to both sides to get all the x terms on one side:\[ 2x + 5x + 9 = 0 \]This simplifies to:\[ 7x + 9 = 0 \]

Step 3: Solve for x

Subtract 9 from both sides:\[ 7x = -9 \]Now, divide by 7:\[ x = -\frac{9}{7} \]

Step 4: Check the Domain

Check if the solution \(x = -\frac{9}{7}\) makes the arguments of the logarithms positive. For \(\ln(2x + 9)\) and \(\ln(-5x)\):- \(2(-\frac{9}{7}) + 9 = \frac{9}{7} > 0\) is OK.- \(-5(-\frac{9}{7}) = \frac{45}{7} > 0\) is OK.

Step 5: Graph Both Sides of the Equation

Graph the functions \(y = \ln(2x + 9)\) and \(y = \ln(-5x)\). The solution should be the x-coordinate of their intersection point.

Key Concepts

Logarithmic Equations

Logarithmic equations involve variables located inside a logarithm. To solve them, a common approach is to first express both sides of the equation as logarithms with the same base. For example, consider the equation given:

• \( \ln (2x+9) = \ln (-5x) \)

Since both sides have the natural logarithm (\( \ln \)) with base \( e \), you can equate the arguments of the logarithms directly. This property markedly simplifies the equation since it shifts focus from the logarithms to just the expressions within.
By setting the arguments equal, you solve the expression:

• \( 2x + 9 = -5x \)

This simplification is fundamental to tackling and solving logarithmic equations effectively.

Graphing Functions

Graphing functions visually represents equations, giving a clear insight into their behavior. To graph the equation \( \ln (2x+9) \) and \( \ln (-5x) \), consider each side as a separate function:

• Function 1: \( y = \ln(2x + 9) \)
• Function 2: \( y = \ln(-5x) \)

These functions can be graphed over a suitable range of \( x \) values. Pay attention to the domain of the functions to ensure only valid (positive) numbers are used for the input.
Graphing provides a visual method to verify solutions—by observing where these two graphs intersect on the Cartesian plane, if at all. This visual cross-check helps in confirming solutions won algebraically.

Intersection Point

The intersection point of two graphs represents a common solution to the corresponding equations. In the context of solving \( \ln (2x+9) = \ln (-5x) \), the solution found algebraically \( x = -\frac{9}{7} \) becomes the \( x \)-coordinate of the intersection.
Finding the intersection:

• Graph both \( y = \ln(2x+9) \) and \( y = \ln(-5x) \).
• The intersection point is \( x = -\frac{9}{7} \).

This confirms that the solution meets the criterion for both functions, ensuring they're equal at this value of \( x \). Visually validating this point on the graph solidifies understanding, linking algebraic techniques to geometrical representations.

Domain Verification

Domain verification is crucial to ensure that all values used in the equations are valid and meaningful. In the case of logarithmic equations like \( \ln (2x+9) = \ln (-5x) \), we must inspect whether the arguments maintain positivity:

• For \( \ln(2x + 9) \), ensure \( 2x + 9 > 0 \). Check \( 2(-\frac{9}{7}) + 9 = \frac{9}{7} > 0 \).
• For \( \ln(-5x) \), ensure \( -5x > 0 \). Check \(-5(-\frac{9}{7}) = \frac{45}{7} > 0 \).

Both conditions are satisfied, indicating \( x = -\frac{9}{7} \) is indeed a valid solution. This verification step is essential, as it confirms the real and permissible domain values where the logarithmic function is defined, avoiding unexpected errors.