Question

Iog5(x^2-4)-log5(x-2)=0

Answer 1

Дано уравнение Iog5(x^2-4)-log5(x-2)=0.

По свойству разности логарифмов при одинаковом основании имеем:

Iog5((x^2-4/(x-2))=0 при х ≠ 2.

Так как 5^0 = 1, то (x^2-4/(x-2)) = 1.

Или x^2-4 = x-2.

Имеем квадратное уравнение x^2 – x – 2 = 0.

D = 1-4*1*(-2) = 9. √D = +-3.

x1 = (1 – 3)/2 = -1.

X2 = (1 + 3)/2 = 2, но это решение не проходит по ОДЗ .

Ответ: х = -1.

Answer 2

To solve the equation log₅(x² - 4) - log₅(x - 2) = 0, we can use logarithmic properties to simplify it.

First, recall the logarithmic property: logₐ(b) - logₐ(c) = logₐ(b/c).

Applying this property to the given equation, we have: log₅((x² - 4)/(x - 2)) = 0.

Next, we can rewrite the equation in exponential form: 5^0 = (x² - 4)/(x - 2).

Since any number raised to the power of 0 is equal to 1, we have: 1 = (x² - 4)/(x - 2).

Now, we can cross-multiply to get rid of the fraction: x² - 4 = x - 2.

Rearranging the equation, we have: x² - x - 2 = 4.

Bringing all terms to one side, we get: x² - x - 6 = 0.

Now, we can factor the quadratic equation: (x - 3)(x + 2) = 0.

Setting each factor to zero and solving for x, we have two possible solutions: x - 3 = 0, which gives x = 3. x + 2 = 0, which gives x = -2.

Therefore, the solutions to the equation log₅(x² - 4) - log₅(x - 2) = 0 are x = 3 and x = -2.