a student solved the equation below by graphing.\nlog₆(x - 1) = log₂(2x + 2)\nwhich statement about the…

Explanation:

Step1: Analyze the domain

For (y = \log_{6}(x - 1)), the domain is (x-1>0), i.e., (x>1). For (y=\log_{2}(2x + 2)), the domain is (2x+2>0), i.e., (x>- 1). The common - domain for the two functions is (x>1).

Step2: Consider the growth rates

The function (y = \log_{a}x=\frac{\ln x}{\ln a}). For (y_1=\log_{6}(x - 1)=\frac{\ln(x - 1)}{\ln6}) and (y_2=\log_{2}(2x + 2)=\frac{\ln(2x + 2)}{\ln2}). Take the derivative of (y_1) with respect to (x): (y_1^\prime=\frac{1}{(x - 1)\ln6}). Take the derivative of (y_2) with respect to (x): (y_2^\prime=\frac{2}{(2x + 2)\ln2}=\frac{1}{(x + 1)\ln2}). When (x>1), (\ln6>\ln2>0) and (x + 1>x - 1>0). So (y_1^\prime<y_2^\prime). As (x = 2): (y_1=\log_{6}(2 - 1)=\log_{6}1 = 0), (y_2=\log_{2}(2\times2+2)=\log_{2}6>0). As (x\to+\infty), (\lim_{x\to+\infty}\log_{6}(x - 1)=\lim_{x\to+\infty}\frac{\ln(x - 1)}{\ln6}) and (\lim_{x\to+\infty}\log_{2}(2x + 2)=\lim_{x\to+\infty}\frac{\ln(2x + 2)}{\ln2}). Using L'Hopital's rule (for (\frac{\infty}{\infty}) form (\lim_{x\to+\infty}\frac{f(x)}{g(x)}=\lim_{x\to+\infty}\frac{f^\prime(x)}{g^\prime(x)})), (\lim_{x\to+\infty}\log_{6}(x - 1)=\lim_{x\to+\infty}\frac{1}{x - 1}\times\frac{1}{\ln6}=0) and (\lim_{x\to+\infty}\log_{2}(2x + 2)=\lim_{x\to+\infty}\frac{2}{2x + 2}\times\frac{1}{\ln2}=0). But (y_2) grows faster than (y_1) in the common - domain (x>1).

We can also use the fact that if we rewrite the equations as (y_1=\log_{6}(x - 1)) and (y_2=\log_{2}(2x + 2)). Let (f(x)=\log_{6}(x - 1)-\log_{2}(2x + 2)=\frac{\ln(x - 1)}{\ln6}-\frac{\ln(2x + 2)}{\ln2}). (f^\prime(x)=\frac{1}{(x - 1)\ln6}-\frac{2}{(2x + 2)\ln2}=\frac{1}{(x - 1)\ln6}-\frac{1}{(x + 1)\ln2}). Since (\ln6>\ln2) and (x+1>x - 1) for (x > 1), (f^\prime(x)<0) for (x>1). (f(x)) is a decreasing function on ((1,+\infty)).

Answer:

The curves do not intersect.