Find the number of real values of x for which log 2x = 2 log (x + 1) is satisfied.

Sure, let's solve the equation step by step.

1. First, we simplify the equation using the properties of logarithms. The equation log 2x = 2 log (x + 1) can be rewritten as log 2x = log (x + 1)^2.

2. Now, we can remove the logarithms from both sides of the equation because if log a = log b, then a = b. So, we get 2x = (x + 1)^2.

3. Expanding the right side of the equation, we get 2x = x^2 + 2x + 1.

4. Rearranging the terms, we get x^2 + 2x - 2x - 1 = 0, which simplifies to x^2 - 1 = 0.

5. This equation can be factored as (x - 1)(x + 1) = 0.

6. Setting each factor equal to zero gives the solutions x = 1 and x = -1.

7. However, we must check these solutions in the original equation because logarithms are only defined for positive numbers.

8. Substituting x = 1 into the original equation, we get log 2(1) = 2 log (1 + 1), which simplifies to log 2 = log 2. So, x = 1 is a solution.

9. Substituting x = -1 into the original equation, we get log 2(-1) = 2 log (-1 + 1), which is undefined because we can't take the logarithm of a negative number or zero. So, x = -1 is not a solution.

Therefore, there is only one real value of x, which is x = 1, for which the equation log 2x = 2 log (x + 1) is satisfied.

Sure, let's solve the equation step by step.

1. We have the equation log 2x = 2 log (x + 1).

2. We can rewrite the right side of the equation using the property of logarithms that says log(a^b) = b log a. So, 2 log (x + 1) becomes log (x + 1)^2.

3. Now our equation is log 2x = log (x + 1)^2.

4. Since the logarithms on both sides of the equation have the same base, we can set the arguments equal to each other. So, 2x = (x + 1)^2.

5. Expanding the right side of the equation gives us 2x = x^2 + 2x + 1.

6. Rearranging the terms gives us x^2 - 2x + 1 = 0.

7. This is a quadratic equation in the form ax^2 + bx + c = 0, where a = 1, b = -2, and c = 1. We can solve for x using the quadratic formula x = [-b ± sqrt(b^2 - 4ac)] / (2a).

8. Substituting the values of a, b, and c into the quadratic formula gives us x = [2 ± sqrt((-2)^2 - 411)] / (2*1) = [2 ± sqrt(4 - 4)] / 2 = [2 ± 0] / 2.

9. So, the solutions are x = 1 and x = 1, which are actually the same solution.

10. Therefore, there is only one real value of x that satisfies the equation log 2x = 2 log (x + 1), and that value is x = 1.