Solve Log(x^2 + 4) = 2 + log x -log 20

Answer 1

We can start by using the properties of logarithms to simplify the equation.

First, we can combine the logs on the right side:
log(x^2 + 4) = log(x) + log(20/1)
log(x^2 + 4) = log(20x)

Next, we can use the fact that log b + log c = log (b * c):
log(x^2 + 4) = log(20x)

By setting the expressions inside the logs equal to each other, we get:
x^2 + 4 = 20x

Rearranging the equation:
x^2 - 20x + 4 = 0

Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula.

However, upon further inspection, this equation cannot be factored easily. So, we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 1, b = -20, and c = 4:

x = (-(-20) ± √((-20)^2 - 4(1)(4))) / (2(1))
x = (20 ± √(400 - 16)) / 2
x = (20 ± √384) / 2
x = (20 ± 19.60) / 2

Using the positive square root:
x = (20 + 19.60) / 2
x = 39.60 / 2
x = 19.80

So, the solution to the equation Log(x^2 + 4) = 2 + log x - log 20 is x = 19.80.

Answer 2

log(x^2 + 4) = 2 + log x - log 20
x^2+4 = 100 * x / 20
x^2 + 4 = 5x
x^2-5x+4 = 0
(x-1)(x-4) = 0
x = 1 or 4

Answer 3

The solutions to the equation Log(x^2 + 4) = 2 + log x - log 20 are x = 1 or x = 4.