Question

Log2(x+12)=2*log2x. ))))))))

Answer 1

ОДЗ: x+12>0; x>0, значит х>0.
log_2_(x+12)=log_2_x^2, это равносильно: (x+12)=x^2. Решаем квадратное уравнение: x^2-x-12=0. x1=4, x2=-3 - не подходит по ОДЗ.

Answer 2

To solve the equation log2(x+12) = 2*log2(x), we can use logarithmic properties and algebraic manipulation. Here's the step-by-step solution:

1. Start by applying the power rule of logarithms to the right side of the equation: log2(x+12) = log2(x^2)

2. Since the bases are the same, we can remove the logarithms: x + 12 = x^2

3. Rearrange the equation to set it equal to zero: x^2 - x - 12 = 0

4. Factorize the quadratic equation: (x - 4)(x + 3) = 0

5. Set each factor equal to zero and solve for x: x - 4 = 0 --> x = 4 x + 3 = 0 --> x = -3

6. Check the solutions in the original equation: For x = 4: log2(4+12) = 2log2(4) log2(16) = 22 4 = 4 (True)

   For x = -3: log2(-3+12) = 2log2(-3) log2(9) = 2log2(-3) There is no logarithm of a negative number, so x = -3 is not a valid solution.

Therefore, the solution to the equation log2(x+12) = 2*log2(x) is x = 4.