(log2(x%5E2-3)%2Blog0-5(%E2%88%9A3-x))%3D0.jpeg "(x^2+2x+1)(log2(x^2-3)+log0 5(√3-x))=0")
Solving the Equation: (x^2+2x+1)(log2(x^2-3)+log0 5(√3-x))=0
This equation presents a unique challenge due to the combination of polynomial and logarithmic terms. To solve it, we need to utilize the properties of both types of expressions.
Understanding the Equation
Let's break down the equation:
- (x^2+2x+1): This is a quadratic expression that can be factored as (x+1)^2.
- log2(x^2-3): This is a logarithmic expression with base 2. It is defined only when x^2 - 3 > 0.
- log0 5(√3-x): This is a logarithmic expression with base 0.5. It is defined only when √3 - x > 0.
Solving for x
The equation is satisfied when either factor equals zero:
Case 1: (x+1)^2 = 0
This implies x = -1.
Case 2: log2(x^2-3) + log0 5(√3-x) = 0
To solve this part, we need to use logarithmic properties:
- log a + log b = log (a*b): Combining the logarithms.
- log a = log b => a = b: Solving for the argument.
Applying these properties, we get:
log2[(x^2-3)(√3-x)] = 0
This implies:
(x^2-3)(√3-x) = 2^0 = 1
Expanding and simplifying this equation leads to a cubic equation:
x^3 - √3x^2 + 3x - 3√3 - 1 = 0
Solving this cubic equation will require numerical methods or factorization techniques.
Analyzing the Solutions
- x = -1: This solution is valid as it satisfies the conditions for both logarithmic terms.
- Solutions from the cubic equation: We need to ensure that these solutions satisfy the domain restrictions of the logarithmic terms.
Important Note: The cubic equation might have solutions that are not valid due to the restrictions imposed by the logarithms.
Conclusion
Solving the given equation requires a combination of factoring, logarithmic properties, and possibly numerical methods. We need to pay close attention to the domain restrictions of the logarithmic expressions to determine valid solutions.