Question

Log7(x-4)+log7(x+1)=log7(4x+4)

Answer 1

1) log7(x - 4) = 1 + log7(x - 10).

Разберем ОДЗ:

х - 4 > 0; x > 4.

x - 10 > 0; x > 10.

Общее решение ОДЗ: x > 10.

Представим единицу как логарифм с основанием 7:

log7(x - 4) = log77 + log7(x - 10).

По правилу сложения логарифмов:

log7(x - 4) = log7(7 * (x - 10)).

log7(x - 4) = log7(7x - 70).

Избавляемся от логарифмов:

х - 4 = 7х - 70.

Перевернем для удобства:

7х - 70 = х - 4.

7х - х = 70 - 4.

6х = 66.

х = 11 (подходит по ОДЗ).

Ответ: корень уравнения равен 11.

2) log2(x² - 4x + 4) = 4.

ОДЗ: x² - 4x + 4 > 0.

Это квадратичная парабола, ветви вверх. Точка пересечения с осью х: х = 2 (по теореме Виета). Так как ветки параболы вверх и неравенство строгое, то решение ОДЗ: х принадлежит (-∞; 2) и (2; +∞).

Представим 4 как логарифм с основанием 2:

log2(x² - 4x + 4) = log216.

Избавляемся от логарифмов.

x² - 4x + 4 = 16.

x² - 4x - 12 = 0.

Корни уравнения по теореме Виета равны -2 и 6. Оба корня подходят под условие ОДЗ.

Ответ: корни уравнения равны -2 и 6.

Answer 2

To solve the logarithmic equation, we'll use logarithmic properties to combine the logarithms into a single expression. The logarithmic property we'll use is:

log(base a)(b) + log(base a)(c) = log(base a)(b * c)

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

log7(x-4) + log7(x+1) = log7(4x+4)

Applying the property:

log7((x-4)*(x+1)) = log7(4x+4)

Now, according to the one-to-one property of logarithms, if two logarithms with the same base are equal, their arguments must also be equal:

(x-4)*(x+1) = 4x + 4

Let's solve for x:

x^2 - 4x + x - 4 = 4x + 4

Simplify:

x^2 - 3x - 4 = 4x + 4

Now, move all terms to one side of the equation:

x^2 - 3x - 4x - 4 - 4 = 0

x^2 - 7x - 8 = 0

Now, we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

For our equation, a = 1, b = -7, and c = -8:

x = (7 ± √((-7)^2 - 4 * 1 * (-8))) / 2 * 1 x = (7 ± √(49 + 32)) / 2 x = (7 ± √81) / 2 x = (7 ± 9) / 2

Now, we have two potential solutions for x:

1. x = (7 + 9) / 2 x = 16 / 2 x = 8

2. x = (7 - 9) / 2 x = -2 / 2 x = -1

However, we need to check these solutions in the original equation since sometimes they might not satisfy the domain of the logarithmic function.

Let's check x = 8:

log7(8-4) + log7(8+1) = log7(4*8+4) log7(4) + log7(9) = log7(32+4) 1 + log7(9) = log7(36) 1 + 2 = 2 (This is correct)

Now, let's check x = -1:

log7(-1-4) + log7(-1+1) = log7(4*(-1)+4) log7(-5) + log7(0) = log7(-4) The logarithm of a negative number is undefined, and the logarithm of 0 is also undefined.

Therefore, x = -1 is not a valid solution.

So, the only valid solution to the equation is x = 8.