Question

Log6 (4x+55) - log6 (7+x) = Log6 (25+3x)

Answer 1

Log6(4x+55)-Log6(7+x)=Log6(25+3x) по свойству логарифмов:
Log6(4x+55)/(7+x)=Log6(25+3x), отсюда
4x+55/7+x=25+3x , домножим обе части на 7+x
4x+55=175+25x+21x+3x^2
3x^2+42x+120=0, решаем квадратное уравнение, его корни равны x=-5 и x=-9, но область опеределения этого уравнения x>-7, значит нам подходит только x=-5.
Ответ: x=-5

Answer 2

To solve the equation involving logarithms, we'll use logarithmic properties and algebraic manipulations. Let's simplify the equation step by step:

1. Start with the given equation: log6(4x + 55) - log6(7 + x) = log6(25 + 3x)

2. Apply the quotient rule of logarithms, which states that log(base a)(b) - log(base a)(c) = log(base a)(b/c): log6((4x + 55) / (7 + x)) = log6(25 + 3x)

3. Remove the logarithm notation by equating the expressions inside the logarithms: (4x + 55) / (7 + x) = 25 + 3x

4. Multiply both sides of the equation by the common denominator (7 + x) to eliminate the fraction: (4x + 55) = (25 + 3x) * (7 + x)

5. Expand the right side of the equation using the distributive property: 4x + 55 = (25 * 7) + (25 * x) + (3x * 7) + (3x * x)

6. Simplify the equation further: 4x + 55 = 175 + 25x + 21x + 3x^2

7. Rearrange the equation to bring all terms to one side: 3x^2 + 21x + 25x - 4x - 175 - 55 = 0

8. Combine like terms: 3x^2 + 42x - 230 = 0

9. To solve this quadratic equation, you can either factorize it or use the quadratic formula. Let's use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

   In our equation, a = 3, b = 42, and c = -230.

   Substituting these values into the quadratic formula: x = (-42 ± √(42^2 - 4 * 3 * -230)) / (2 * 3)

10. Simplify the expression inside the square root: x = (-42 ± √(1764 + 2760)) / 6 x = (-42 ± √(4524)) / 6 x = (-42 ± 67.32) / 6

11. Solve for both possible values of x: x1 = (-42 + 67.32) / 6 ≈ 4.22 x2 = (-42 - 67.32) / 6 ≈ -18.22

Therefore, the equation log6(4x + 55) - log6(7 + x) = log6(25 + 3x) has two solutions: x ≈ 4.22 and x ≈ -18.22.