What is the value of x in the logarithmic equation log(x + 21) + log x = 2?

To solve the equation log(x + 21) + log x = 2, we can use the properties of logarithms. One useful property is that the sum of two logarithms is equal to the logarithm of the product of their arguments. Thus, we can combine the logarithmic terms:

log((x + 21)x) = 2

This simplifies to:

log(x^2 + 21x) = 2

Next, we can remove the logarithm by converting to its exponential form. This gives us:

x^2 + 21x = 10^2

Since 10^2 equals 100, we have:

x^2 + 21x - 100 = 0

Now, we need to solve this quadratic equation using the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

Here, a = 1, b = 21, and c = -100. Plugging these values into the formula:

x = (-21 ± √(21^2 - 4(1)(-100))) / (2 * 1)

x = (-21 ± √(441 + 400)) / 2

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