Question

a student solved the equation below by graphing.

\\(\log_{6}(x-1) = \log_{2}(2x+2)\\)

which statement about the graph is true?

• the curves do not intersect.
• the curves intersect at one point.
• the curves intersect at two points.
• the curves appear to coincide.

Explanation:

Analyze the domains of the logarithmic functions

Using the Logarithmic Functions knowledge point
The domain of \(y = \log_6(x-1)\) requires:

$$ x - 1 > 0 \implies x > 1 $$

The domain of \(y = \log_2(2x+2)\) requires:

$$ 2x + 2 > 0 \implies x > -1 $$

Thus, any intersection must occur in the common domain \(x > 1\).

Evaluate function values to determine intersection behavior

Using the Solving Exponential Equations Graphically knowledge point
Let \(f(x) = \log_6(x-1)\) and \(g(x) = \log_2(2x+2)\).
For \(x > 1\):

• As \(x \to 1^+\), \(f(x) \to -\infty\), while \(g(x) \to \log_2(4) = 2\). Thus, \(f(x) < g(x)\) near \(x = 1\).
• Let's compare values at a larger point, say \(x = 7\):

$$ f(7) = \log_6(6) = 1 $$

$$ g(7) = \log_2(16) = 4 $$

Here, \(f(7) < g(7)\).

• Since both functions are strictly increasing, let's analyze their rates of growth. The derivative of each function is:

$$ f'(x) = \frac{1}{(x-1)\ln(6)} $$

$$ g'(x) = \frac{2}{(2x+2)\ln(2)} = \frac{1}{(x+1)\ln(2)} $$

For all \(x > 1\), we have \(x - 1 < x + 1\) and \(\ln(2) < \ln(6)\). However, because the base of \(g(x)\) is \(2\) (which is much smaller than \(6\)), \(g(x)\) grows significantly faster than \(f(x)\). Specifically, \(g(x)\) starts at a higher value at the boundary \(x=1\) and has a larger value at all evaluated points. Since \(g(x) > f(x)\) for all \(x > 1\), the two curves do not intersect.

Answer:

• The curves do not intersect. (Correct answer)
• The curves intersect at one point.
• The curves intersect at two points.
• The curves appear to coincide.