Question

a student solved the equation below by graphing.\\(\log_{6}(x - 1) = \log_{2}(2x + 2)\\)\
which statement about the graph is true?\
\\(\bigcirc\\) the curves do not intersect.\
\\(\bigcirc\\) the curves intersect at one point.\
\\(\bigcirc\\) the curves intersect at two points.\
\\(\bigcirc\\) the curves appear to coincide.

Explanation:

Step1: Analyze the functions

We have two logarithmic functions: \( y = \log_{6}(x - 1) \) and \( y=\log_{2}(2x + 2) \). The domain of \( y=\log_{6}(x - 1) \) requires \( x-1>0\Rightarrow x > 1 \). The domain of \( y = \log_{2}(2x + 2) \) requires \( 2x+2>0\Rightarrow x>- 1 \). So the common domain for both functions (where we can look for intersections) is \( x>1 \).

Step2: Analyze the behavior of the functions

- For \( y=\log_{6}(x - 1) \), as \( x ightarrow1^{+} \), \( y ightarrow-\infty \), and as \( x ightarrow\infty \), \( y ightarrow\infty \) (since the base \( 6>1 \), the function is increasing).
- For \( y = \log_{2}(2x + 2)=\log_{2}(2(x + 1))=1+\log_{2}(x + 1) \). As \( x ightarrow1^{+} \), \( y = 1+\log_{2}(2)=1 + 1=2 \). As \( x ightarrow\infty \), \( y ightarrow\infty \) (since the base \( 2>1 \), the function is increasing). Also, the derivative of \( y=\log_{6}(x - 1)=\frac{1}{(x - 1)\ln6} \) and the derivative of \( y=\log_{2}(2x + 2)=\frac{2}{(2x + 2)\ln2}=\frac{1}{(x + 1)\ln2} \). We can check the growth rates. At \( x = 2 \):
- \( y_1=\log_{6}(2 - 1)=\log_{6}(1) = 0 \)
- \( y_2=\log_{2}(2\times2+2)=\log_{2}(6)\approx2.58 \)
At \( x = 5 \):
- \( y_1=\log_{6}(5 - 1)=\log_{6}(4)\approx0.861 \)
- \( y_2=\log_{2}(2\times5+2)=\log_{2}(12)\approx3.58 \)
At \( x = 10 \):
- \( y_1=\log_{6}(10 - 1)=\log_{6}(9)\approx1.285 \)
- \( y_2=\log_{2}(2\times10+2)=\log_{2}(22)\approx4.459 \)
We can see that \( \log_{2}(2x + 2) \) is always greater than \( \log_{6}(x - 1) \) for \( x>1 \) (since at \( x = 2 \), \( \log_{2}(6)\approx2.58>0 \); and the function \( \log_{2}(2x + 2) \) has a higher initial value at \( x = 1^{+} \) and both are increasing, but \( \log_{2}(2x + 2) \) grows faster? Wait, no, let's check the derivatives. The derivative of \( \log_{6}(x - 1) \) is \( \frac{1}{(x - 1)\ln6} \) and the derivative of \( \log_{2}(2x + 2) \) is \( \frac{1}{(x + 1)\ln2} \). Let's compare the derivatives at \( x = 2 \): \( \frac{1}{(2 - 1)\ln6}=\frac{1}{\ln6}\approx0.55 \), \( \frac{1}{(2+1)\ln2}=\frac{1}{3\ln2}\approx0.48 \). At \( x = 10 \): \( \frac{1}{(10 - 1)\ln6}=\frac{1}{9\ln6}\approx0.061 \), \( \frac{1}{(10 + 1)\ln2}=\frac{1}{11\ln2}\approx0.13 \). Wait, so the derivative of \( \log_{2}(2x + 2) \) becomes larger than the derivative of \( \log_{6}(x - 1) \) as \( x \) increases. But initially, at \( x = 2 \), \( \log_{6}(1)=0 \) and \( \log_{2}(6)\approx2.58 \). Let's check if there is any intersection. Suppose \( \log_{6}(x - 1)=\log_{2}(2x + 2) \). Let's test \( x = 2 \): LHS = 0, RHS=\( \log_{2}(6)\approx2.58 \), LHS < RHS. \( x = 3 \): LHS=\( \log_{6}(2)\approx0.387 \), RHS=\( \log_{2}(8)=3 \), LHS < RHS. \( x = 4 \): LHS=\( \log_{6}(3)\approx0.613 \), RHS=\( \log_{2}(10)\approx3.32 \), LHS < RHS. As \( x \) increases, \( \log_{6}(x - 1) \) grows slower than \( \log_{2}(2x + 2) \) in the long run? Wait, no, the derivative of \( \log_{6}(x - 1) \) is \( \frac{1}{(x - 1)\ln6} \) (decreasing as \( x \) increases) and the derivative of \( \log_{2}(2x + 2) \) is \( \frac{1}{(x + 1)\ln2} \) (also decreasing as \( x \) increases). But at \( x = 1 \), the first function is not defined (approaches -infty) and the second function at \( x = 1 \) is \( 1+\log_{2}(2)=2 \). Since the first function starts from -infty and the second function at \( x = 1^{+} \) is 2, and both are increasing, but the first function has to catch up from -infty. Wait, when \( x \) is just slightly greater than 1, say \( x = 1.1 \), \( \log_{6}(0.1)\approx - 1.098 \), and \( \log_{2}(2\times1.1 + 2)=\log_{2}(4.2)\approx2.07 \). So the first function is below the second function. As \( x \) increases, the first function goes from -infty to +infty, and the second function goes from 2 to +infty. Since the first function is increasing and starts below the second function, and both go to +infty, we need to check if the first function will ever cross the second function. Let's consider the limit as \( x ightarrow\infty \). We know that for logarithmic functions, \( \log_{a}x \) and \( \log_{b}x \) (with \( a,b>1 \)) have the same growth rate (they are asymptotic to each other in terms of growth rate, but with different constants). But here we have \( \log_{6}(x - 1) \) and \( \log_{2}(2x + 2) \). Let's rewrite \( \log_{2}(2x + 2)=\log_{2}(2)+\log_{2}(x + 1)=1+\log_{2}(x + 1) \). And \( \log_{6}(x - 1)=\frac{\ln(x - 1)}{\ln6} \), \( \log_{2}(x + 1)=\frac{\ln(x + 1)}{\ln2} \). So we have the equation \( \frac{\ln(x - 1)}{\ln6}=1+\frac{\ln(x + 1)}{\ln2} \). Rearranging: \( \ln(x - 1)=\ln6+\frac{\ln6}{\ln2}\ln(x + 1) \). Since \( \frac{\ln6}{\ln2}=\log_{2}6>2 \), the right - hand side grows faster than the left - hand side (because the coefficient of \( \ln(x + 1) \) is greater than 1 and we also have a constant term). So the left - hand side (which is \( \ln(x - 1) \)) will never catch up to the right - hand side. So the two curves do not intersect.

Answer:

The curves do not intersect.