Question

solve \\(\log(x + 1) = -x^2 + 10\\) by graphing

which equations should be graphed?
\\(y_1 = -x^2\\)
\\(y_1 = -x^2 + 10\\)
\\(y_2 = \frac{\log x}{\log 1}\\)
\\(y_2 = \log(x + 1)\\)

graph the system. use the graph to approximate the solution of the equation
\\(\log(x + 1) = -x^2 + 10\\).

\\(x \approx\\)

Explanation:

Identify the system of equations

Using the Graphical Solution of Equations knowledge point
To solve the equation \(\log(x + 1) = -x^2 + 10\) by graphing, we split the equation into a system of two equations by setting each side equal to a separate dependent variable:

$$
LATEXBLOCK0
$$

Find the intersection point

Using the Solving Exponential Equations Graphically knowledge point
We find the \(x\)-value where the graphs of \(y_1 = -x^2 + 10\) and \(y_2 = \log(x + 1)\) intersect.

• For \(x \approx 3\):

$$
LATEXBLOCK1
$$

• For \(x \approx 3.1\):

$$
LATEXBLOCK2
$$

• Setting \(y_1 = y_2\):

$$ -x^2 + 10 = \log_{10}(x + 1) \implies x \approx 3.05 $$

Thus, the approximate solution is \(x \approx 3\).

Answer:

Question 1

Which equations should be graphed?

• \(y_1 = -x^2\)
• \(y_1 = -x^2 + 10\) (Correct answer)
• \(y_2 = \frac{\log x}{\log 1}\)
• \(y_2 = \log(x + 1)\) (Correct answer)

Question 2

Graph the system. Use the graph to approximate the solution of the equation \(\log(x + 1) = -x^2 + 10\).
<blank>\(x \approx 3\)</blank>