To determine the number of intersection points between the curves represented by the equations [tex]\(\log_6(x-1)\)[/tex] and [tex]\(\log_2(2x+2)\)[/tex], we need to solve the equation:
[tex]\[
\log_6(x-1) = \log_2(2x+2)
\][/tex]
Let's break this down step-by-step.
1. Understand the Domain:
- [tex]\(\log_6(x-1)\)[/tex] is defined when [tex]\(x-1 > 0\)[/tex], hence:
[tex]\[
x > 1
\][/tex]
- [tex]\(\log_2(2x+2)\)[/tex] is defined when [tex]\(2x+2 > 0\)[/tex], hence:
[tex]\[
x > -1
\][/tex]
The intersection of these domains is [tex]\(x > 1\)[/tex].
2. Equate the Two Logs:
Rewriting the equation using the change of base formula:
[tex]\[
\frac{\log(x-1)}{\log(6)} = \frac{\log(2x+2)}{\log(2)}
\][/tex]
This simplifies to:
[tex]\[
\log_2(2x+2) = \log_2(2(x+1))
\][/tex]
3. Equate the Arguments of the Logs:
Since [tex]\(\log_a(b) = \log_a(c)\)[/tex] if and only if [tex]\(b = c\)[/tex], we get:
[tex]\[
x-1 = 2^{\log_2(2(x+1))}
\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[
x-1 = 2(x+1)
\][/tex]
Solving this equation:
[tex]\[
x - 1 = 2x + 2
\][/tex]
[tex]\[
-3 = x
\][/tex]
This solution is not valid because [tex]\(-3\)[/tex] is not within the domain [tex]\(x > 1\)[/tex].
4. Look for Possible Other Solutions:
Let's check further by plotting or analyzing the behavior of both functions.
- [tex]\(\log_6(x-1)\)[/tex] increases logarithmically starting from [tex]\(x=1\)[/tex].
- [tex]\(\log_2(2x+2)\)[/tex] also increases logarithmically but grows faster than [tex]\(\log_6(x-1)\)[/tex] since the base of the logarithm [tex]\(2\)[/tex] is smaller than [tex]\(6\)[/tex].
Since the only algebraic solution (-3) does not lie in the valid domain [tex]\(x > 1\)[/tex], there are no valid solutions in the domain.
Thus, we conclude:
[tex]\[
\text{The curves do not intersect.}
\][/tex]
Hence, the correct statement about the graph is:
- The curves do not intersect.
