Question

Lg(x^2+100x)-lg(x+4)=2​

Answer 1

Ответ:

Пошаговое объяснение:

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Answer 2

To solve the equation lg(x^2 + 100x) - lg(x + 4) = 2, we need to apply some logarithmic properties and algebraic manipulations. First, recall the logarithmic property:

lg(a) - lg(b) = lg(a/b)

Using this property, we can rewrite the equation as follows:

lg((x^2 + 100x)/(x + 4)) = 2

Now, let's eliminate the logarithm by converting it into an exponential form. In general, if lg(c) = d, then c = 10^d. Applying this, we get:

(x^2 + 100x)/(x + 4) = 10^2

Simplify the right-hand side:

(x^2 + 100x)/(x + 4) = 100

Next, we can get rid of the fraction by multiplying both sides by (x + 4):

x^2 + 100x = 100(x + 4)

Now, expand the right-hand side:

x^2 + 100x = 100x + 400

Subtract 100x from both sides:

x^2 = 400

Now, take the square root of both sides:

x = ±√400

x = ±20

So, the solutions to the equation are x = 20 and x = -20. However, we need to verify that these solutions are valid because we're dealing with logarithms.

For the original equation, the argument inside the logarithm should always be positive. Hence, x cannot be -20 because it would make (x + 4) in the denominator negative.

Therefore, the only valid solution is x = 20. Let's check it:

lg(20^2 + 100*20) - lg(20 + 4) = lg(400 + 2000) - lg(24) = lg(2400) - lg(24)

Now, we can calculate:

lg(2400) ≈ 3.3802 lg(24) ≈ 1.3802

So, the equation becomes:

3.3802 - 1.3802 ≈ 2 (approximately)

Thus, x = 20 is the valid solution to the equation.