A student solved the equation below by graphing.
[tex]$\log _6(x-1)=\log _2(2 x+2)$[/tex]
Which statement about the graph is true?
A. The curves do not intersect.
B. The curves intersect at one point.
C. The curves intersect at two points.
D. The curves appear to coincide.
Answer (1)
Determine the domain of both logarithmic functions, which is 1"> x > 1 .
Analyze the behavior of the functions as x approaches 1 from the right and as x approaches infinity.
Find the derivatives of both functions and determine where they are equal.
Conclude that the curves intersect at one point since the derivatives are equal at only one point and the functions approach different values as x approaches 1. T h ec u r v es in t ersec t a t o n e p o in t .
Explanation
Understanding the Problem We are given the equation lo g 6 ( x − 1 ) = lo g 2 ( 2 x + 2 ) and asked to determine the number of intersection points of the graphs of y = lo g 6 ( x − 1 ) and y = lo g 2 ( 2 x + 2 ) .
Finding the Domain First, we need to find the domain of each function. The domain of lo g 6 ( x − 1 ) is 0"> x − 1 > 0 , so 1"> x > 1 . The domain of lo g 2 ( 2 x + 2 ) is 0"> 2 x + 2 > 0 , so -1"> x > − 1 . Therefore, the domain of the equation is 1"> x > 1 .
Analyzing Behavior Near x=1 Let's analyze the behavior of the functions as x approaches 1 from the right. As x → 1 + , lo g 6 ( x − 1 ) → − ∞ . As x → 1 + , lo g 2 ( 2 x + 2 ) → lo g 2 ( 4 ) = 2 . Since one function approaches negative infinity and the other approaches 2, there must be an intersection point.
Analyzing Behavior as x Approaches Infinity Let's analyze the behavior of the functions as x approaches infinity. As x → ∞ , both functions approach infinity.
Finding the Derivatives To determine if the functions intersect more than once, we can analyze their derivatives. The derivative of lo g 6 ( x − 1 ) is ( x − 1 ) l n ( 6 ) 1 . The derivative of lo g 2 ( 2 x + 2 ) is ( 2 x + 2 ) l n ( 2 ) 2 = ( x + 1 ) l n ( 2 ) 1 .
Comparing the Derivatives Now, let's compare the derivatives to see if they are equal at any point. ( x − 1 ) l n ( 6 ) 1 = ( x + 1 ) l n ( 2 ) 1 . This implies ( x + 1 ) ln ( 2 ) = ( x − 1 ) ln ( 6 ) . Expanding, we get x ln ( 2 ) + ln ( 2 ) = x ln ( 6 ) − ln ( 6 ) . Rearranging, we have x ( ln ( 6 ) − ln ( 2 )) = ln ( 6 ) + ln ( 2 ) . Thus, x ln ( 3 ) = ln ( 12 ) , so x = l n ( 3 ) l n ( 12 ) = l n ( 3 ) l n ( 3 ) + l n ( 4 ) = 1 + l n ( 3 ) l n ( 4 ) = 1 + lo g 3 ( 4 ) ≈ 1 + 1.26 = 2.26 . Since the derivatives are equal at only one point, the functions intersect at most twice. However, since the derivatives are not equal for all x, the functions do not coincide.
Conclusion Since we know the functions intersect at least once, and the derivatives are equal at only one point, the functions intersect at one point. Therefore, the curves intersect at one point.
Examples
Consider modeling the growth of two different populations, one growing at a rate proportional to lo g 6 ( x − 1 ) and another at a rate of lo g 2 ( 2 x + 2 ) . Finding the intersection point of these logarithmic functions helps determine when the growth rates of the two populations are equal. This is useful in ecological studies, resource management, or even in comparing the performance of different investment strategies where growth is modeled logarithmically. Understanding the intersection points provides critical insights into the relative behavior of these systems.
