Certainly! Let's solve the given logarithmic equation step by step using the One-to-One Property:
1. Given Equation:
[tex]\[\log(x^2 + 3x) = \log(28)\][/tex]
2. Apply the One-to-One Property:
The One-to-One Property of logarithms states that if [tex]\(\log(a) = \log(b)\)[/tex], then [tex]\(a = b\)[/tex]. Applying this property:
[tex]\[x^2 + 3x = 28\][/tex]
3. Form the Quadratic Equation:
Move everything to one side to set the equation to zero:
[tex]\[x^2 + 3x - 28 = 0\][/tex]
4. Identify the Coefficients:
This is a standard quadratic equation in the form [tex]\(ax^2 + bx + c = 0\)[/tex] where:
- [tex]\(a = 1\)[/tex]
- [tex]\(b = 3\)[/tex]
- [tex]\(c = -28\)[/tex]
5. Calculate the Discriminant:
The discriminant [tex]\(\Delta\)[/tex] is given by [tex]\(b^2 - 4ac\)[/tex]:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substitute the values for [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[ \Delta = 3^2 - 4 \cdot 1 \cdot (-28) \][/tex]
[tex]\[ \Delta = 9 + 112 = 121 \][/tex]
6. Solve the Quadratic Equation:
Use the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex]:
[tex]\[ x = \frac{-3 \pm \sqrt{121}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm 11}{2} \][/tex]
This yields two possible solutions:
[tex]\[ x = \frac{-3 + 11}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x = \frac{-3 - 11}{2} = \frac{-14}{2} = -7 \][/tex]
7. Conclusion:
The solutions for the equation [tex]\(\log(x^2 + 3x) = \log(28)\)[/tex] are:
[tex]\[ \boxed{x = 4 \text{ or } x = -7} \][/tex]
