Problem 346 - Precalculus

Question

Solve the equation for \(x,\) if there is a solution. Then graph both sides of the equation, and observe the point of intersection (if it exists) to verify the solution. \(-7+\log _{3}(4-x)=-6\)

Step-by-Step Solution

Answer

The solution to the equation is \(x = 1\).

Step 1: Isolate the Logarithmic Expression

Start by isolating the logarithmic expression on one side of the equation. Given the equation is \[-7 + \log_{3}(4-x) = -6\]Add 7 to both sides to achieve:\[\log_{3}(4-x) = 1\]

Step 2: Convert Logarithmic Equation to Exponential Form

The equation from Step 1 is \(\log_{3}(4-x) = 1\). To solve for \(x\), convert this into its exponential form:\[4-x = 3^1\]This simplifies to:\[4-x = 3\]

Step 3: Solve the Equation for x

From Step 2, solve for \(x\):\[4-x = 3\]Subtract 3 from both sides:\[4-3 = x\]Thus, \(x = 1\).

Step 4: Graph Both Sides of the Original Equation

Graph the functions \(y = -7 + \log_{3}(4-x)\) and \(y = -6\). These graphs help you see where the solution occurs visually, which is at the point of intersection between the two. Since \(x = 1\), in your graph you should observe that both functions intersect at \((1, -6)\).

Step 5: Verify the Solution Matches the Intersection

Verify that when \(x = 1\), both sides of the equation equal the point on the graph:- Calculate \(-7 + \log_{3}(4-1)\) which gives \(-7 + 1 = -6\).This confirms that the graphs coincide at the point \((1, -6)\), verifying that \(x = 1\) is indeed the solution.

Key Concepts

Graphical Solution

A graphical solution refers to using a graph to find where two equations or functions intersect. By graphing each side of the equation separately, you can visually identify the point where they both meet. In our exercise, we deal with two functions:

• Function 1: \( y = -7 + \log_{3}(4-x) \)
• Function 2: \( y = -6 \)

These equations represent a logarithmic curve and a horizontal line, respectively. By plotting both on the same graph, we can look for their intersection point, which confirms the solution found algebraically. When graphed, these functions intersect at the point \((1, -6)\), illustrating that the solution \(x = 1\) is valid.

Exponential Form

Converting logarithmic equations into exponential form can simplify solving for unknown variables. The expression \(\log_{3}(4-x) = 1\) can be rewritten to reveal the base, the result, and the exponent:

• Base: 3
• Exponent: 1
• Result: \(4-x\)

This conversion tells us that \(4-x\) is equal to \(3^1\), or simply 3. So, the equation transforms from a logarithmic one to a straightforward algebraic equation: \(4-x = 3\). Solving this equation simplifies finding the value of \(x\). So remember, the exponential form allows us to see the relationship between the numbers involved more clearly.

Solution Verification

After finding a solution algebraically, it's important to verify that it is correct. This ensures there's no error in the calculation. Once we found \(x = 1\), we plugged it back into the original equation:
\(-7 + \log_{3}(4-1) = -7 + \log_{3}(3)\)
Since \(\log_{3}(3) = 1\), it simplifies to \(-7 + 1 = -6\). Since the solution satisfies the original equation, our answer \(x = 1\) is confirmed correct. This step serves as a check and helps to solidify understanding of the process. If both the algebraic and graphical solutions align with the original equation, it confirms the solution's validity.

Intersection Point

The intersection point is where two graphs meet on the Cartesian plane. In the context of this exercise, the intersection point \((1, -6)\) represents the solution to the equation. This point is both a graphical indication and an algebraic confirmation of the answer. When we graph:

• The logarithmic curve \(y = -7 + \log_{3}(4-x)\)
• The horizontal line \(y = -6\)

The intersection at \((1, -6)\) visually proves the correctness of the solution \(x = 1\). Intersection points are powerful because they provide a visual confirmation of where two conditions are simultaneously satisfied. Always remember: if all steps are correctly performed, the intersection should match the solution derived from any algebraic manipulations.