(b) Solve $2(\log _1 x-1)=\log _3 x$ and $y=\sqrt{x}+1$ simultaneously.
Answer (2)
Correct the flawed equation to 2 ( lo g 2 x − 1 ) = lo g 3 x .
Solve for x by expressing both logarithms in base 2 and substituting z = lo g 2 x , resulting in x ≈ 2.753 .
Substitute the value of x into the second equation y = x + 1 to find y ≈ 2.659 .
The solution to the system of equations is approximately x = 2.753 and y = 2.659 , so the final answer is x = 2.753 , y = 2.659 .
Explanation
Identifying the Flaw and Correcting the Equation The problem asks us to solve a system of equations: 2 ( lo g 1 x − 1 ) = lo g 3 x and y = x + 1 . However, we immediately notice that lo g 1 x is undefined for any x , because the base of a logarithm cannot be 1. This indicates there's likely a typo in the problem statement. We will proceed by correcting the equation to 2 ( lo g 2 x − 1 ) = lo g 3 x and solving the system with the corrected equation.
Rewriting the Corrected Equation Let's correct the first equation to 2 ( lo g 2 x − 1 ) = lo g 3 x . This can be rewritten as 2 lo g 2 x − 2 = lo g 3 x . To solve this, we'll use the change of base formula to express both logarithms in terms of a common base, say base 2. Recall that lo g a b = l o g c a l o g c b . Thus, lo g 3 x = l o g 2 3 l o g 2 x . Substituting this into our equation, we get 2 lo g 2 x − 2 = l o g 2 3 l o g 2 x .
Solving for z Let z = lo g 2 x . Then the equation becomes 2 z − 2 = l o g 2 3 z . Multiplying both sides by lo g 2 3 , we have 2 z lo g 2 3 − 2 lo g 2 3 = z . Rearranging the terms, we get 2 z lo g 2 3 − z = 2 lo g 2 3 , which simplifies to z ( 2 lo g 2 3 − 1 ) = 2 lo g 2 3 . Therefore, z = 2 l o g 2 3 − 1 2 l o g 2 3 .
Solving for x Now we can find x using the relationship z = lo g 2 x . So, x = 2 z = 2 2 l o g 2 3 − 1 2 l o g 2 3 . We can approximate the value of x . First, lo g 2 3 ≈ 1.585 . Then, z = 2 ( 1.585 ) − 1 2 ( 1.585 ) = 3.17 − 1 3.17 = 2.17 3.17 ≈ 1.461 . Therefore, x = 2 1.461 ≈ 2.753 .
Solving for y Finally, we can find y using the equation y = x + 1 . Substituting the value of x we found, y = 2.753 + 1 ≈ 1.659 + 1 = 2.659 . Thus, the solution to the system of equations is approximately x ≈ 2.753 and y ≈ 2.659 .
Final Answer Therefore, the solution to the system of equations 2 ( lo g 2 x − 1 ) = lo g 3 x and y = x + 1 is approximately x = 2.753 and y = 2.659 .
Examples
Imagine you are designing a bridge and need to calculate the tension on a cable. The relationship between the tension and the length of the cable might be expressed using logarithmic and square root functions, similar to the equations we solved. By solving such a system of equations, engineers can determine the precise tension required for the bridge to be stable. This ensures the safety and durability of the structure, preventing potential collapses or failures. Understanding how to solve these equations is crucial for accurate and reliable engineering designs.
To solve the equations, we correct the first equation to 2 ( lo g 2 x − 1 ) = lo g 3 x and find x ≈ 2.753 . Substituting x into y = x + 1 gives y ≈ 2.659 . Thus, the final solutions are x ≈ 2.753 and y ≈ 2.659 .
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