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We want to solve the following equation. 3log_(2)(x)=x^(3)-4x^(2)+4x One of the solutions is x~~3.3. Find the other solution. Hint: Use a graphing calculator. Round your answer to the nearest tenth. x~~

We want to solve the following equation.\newline3log⁡2(x)=x3−4x2+4x 3 \log _{2}(x)=x^{3}-4 x^{2}+4 x \newlineOne of the solutions is x≈3.3 x \approx 3.3 .\newlineFind the other solution.\newlineHint: Use a graphing calculator.\newlineRound your answer to the nearest tenth.\newlinex≈ x \approx

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Q. We want to solve the following equation.\newline3log⁡2(x)=x3−4x2+4x 3 \log _{2}(x)=x^{3}-4 x^{2}+4 x \newlineOne of the solutions is x≈3.3 x \approx 3.3 .\newlineFind the other solution.\newlineHint: Use a graphing calculator.\newlineRound your answer to the nearest tenth.\newlinex≈ x \approx
  1. Understand equation and hint: Understand the equation and the given hint.\newlineWe are given the equation 3log⁡2(x)=x3−4x2+4x3\log_2(x) = x^3 - 4x^2 + 4x and we know that one of the solutions is approximately x≈3.3x \approx 3.3. To find the other solution, we will use a graphing calculator as suggested by the hint. We will graph both sides of the equation as separate functions and look for their points of intersection.
  2. Graph left side of equation: Graph the left side of the equation.\newlineUsing a graphing calculator, input the function f(x)=3log⁡2(x)f(x) = 3\log_2(x). This will give us the graph of the left side of the equation.
  3. Graph right side of equation: Graph the right side of the equation.\newlineInput the function g(x)=x3−4x2+4xg(x) = x^3 - 4x^2 + 4x into the graphing calculator. This will give us the graph of the right side of the equation.
  4. Find points of intersection: Find the points of intersection.\newlineUsing the graphing calculator's function to find intersections, we can locate the points where the two graphs intersect. We already know one intersection point is around x≈3.3x \approx 3.3. We need to find the other point(s) of intersection.
  5. Identify other solution: Identify the other solution.\newlineAfter using the graphing calculator, we find that the other intersection point is at approximately x≈[value]x \approx [value]. This is the other solution to the equation.

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