Question

2log (-x)=1+log (x+4)​

Answer 1

$x = 5 - \sqrt{65}$

ответ

Answer 2

To solve the equation 2log(-x) = 1 + log(x + 4), you can follow these steps:

Step 1: Start by isolating the logarithmic terms on one side of the equation. You can do this by subtracting log(x + 4) from both sides:

2log(-x) - log(x + 4) = 1

Step 2: Use logarithmic properties to combine the two logarithmic terms on the left side of the equation. Specifically, you can use the rule that says log(a) - log(b) = log(a/b):

log((-x)^2 / (x + 4)) = 1

Step 3: Eliminate the logarithm by converting the equation to exponential form. Recall that if log(base b)(x) = y, then b^y = x. In this case, the base is 10 (common logarithm), so we have:

(-x)^2 / (x + 4) = 10^1

Step 4: Simplify and solve for x:

x^2 / (x + 4) = 10

Step 5: Multiply both sides of the equation by (x + 4) to eliminate the fraction:

x^2 = 10(x + 4)

Step 6: Expand and simplify:

x^2 = 10x + 40

Step 7: Move all terms to one side of the equation to set it to zero:

x^2 - 10x - 40 = 0

Step 8: Now, you can solve this quadratic equation. You can use the quadratic formula:

x = (-b ± √(b² - 4ac)) / (2a)

In this case, a = 1, b = -10, and c = -40. Plug these values into the formula:

x = (10 ± √((-10)² - 4(1)(-40))) / (2(1))

x = (10 ± √(100 + 160)) / 2

x = (10 ± √260) / 2

Now, you can simplify further:

x = (10 ± 2√65) / 2

x = 5 ± √65

So, the solutions to the equation 2log(-x) = 1 + log(x + 4) are x = 5 + √65 and x = 5 - √65.