a student solved the equation below by graphing. log₆(x - 1)=log₂(2x + 2) which statement about the graph is…

Explanation:

Step1: Recall logarithmic function properties

The domain of $y = \log_{6}(x - 1)$ is $x-1>0$ i.e. $x > 1$, and the domain of $y=\log_{2}(2x + 2)$ is $2x+2>0$ i.e. $x>- 1$. The common - domain for comparing the two functions is $x>1$. The function $y = \log_{6}(x - 1)$ and $y=\log_{2}(2x + 2)$ are both increasing functions. Let's consider their growth rates. The general form of a logarithmic function is $y=\log_{a}u$, and its derivative is $y'=\frac{1}{u\ln a}$. For $y_1=\log_{6}(x - 1)$, $y_1'=\frac{1}{(x - 1)\ln6}$ (for $x>1$). For $y_2=\log_{2}(2x + 2)=\log_{2}2+\log_{2}(x + 1)=1+\log_{2}(x + 1)$, $y_2'=\frac{1}{(x + 1)\ln2}$ (for $x>-1$). As $x$ increases from $x = 1$ (in the common - domain), we can also analyze the functions by considering some values. When $x = 2$: $y_1=\log_{6}(2 - 1)=\log_{6}1=0$ $y_2=\log_{2}(2\times2 + 2)=\log_{2}6> \log_{2}4 = 2$ As $x$ gets larger, the function $y=\log_{2}(2x + 2)$ grows faster than $y=\log_{6}(x - 1)$ because the base of the second logarithm ($a = 2$) is larger than the base of the first logarithm ($a = 6$) and the argument of the second logarithm ($2x + 2$) is larger than the argument of the first logarithm ($x - 1$) for $x>1$. So the two curves do not intersect.

Answer:

The curves do not intersect.