Question

1) log 3(2x²-5)=2 2) log x-1(2x²-3)=2

Answer 1

Ответ:

+-sqrt7; -1+sqrt5

Объяснение:

1) log3(2x^2-5)=2
2x^2 - 5 = 9
2x^2 = 14
x^2 = 7
x = +-sqrt7

2) log x-1 (2x^2-3) = 2
ОДЗ: x-1 > 0 и не равен 1. (x>1, не равен 2)
2x^2-3 = (x-1)^2
2x^2 - 3 = x^2 - 2x + 1
x^2 + 2x - 4 = 0
D = 4 + 16 = 20
x1, x2 = (-2 +- 2sqrt5)/2 = -1 +- sqrt5.
Корень -1-sqrt5 не подходит по ОДЗ (меньше нуля). Значит, корень - -1+sqrt5.

(sqrt(x) = квадратный корень из x)

Answer 2

To solve these equations, we'll use logarithmic properties to isolate the logarithm expressions and then solve for the variables. Let's tackle each equation step by step:

1. log₃(2x² - 5) = 2

To eliminate the logarithm, we can rewrite the equation in exponential form:

3² = 2x² - 5

Simplifying the left side:

9 = 2x² - 5

Next, isolate the term with x²:

2x² = 9 + 5

2x² = 14

Now divide both sides by 2 to solve for x²:

x² = 14/2

x² = 7

To find the values of x, take the square root of both sides:

x = ±√7

Therefore, the solutions to the equation are x = √7 and x = -√7.

2. logₓ₋₁(2x² - 3) = 2

Following a similar approach, let's convert the equation into exponential form:

(x - 1)² = 2x² - 3

Expanding the left side:

x² - 2x + 1 = 2x² - 3

Rearranging the terms:

2x² - x² - 2x - 3 - 1 = 0

x² - 2x - 4 = 0

Now, we can either factor this quadratic equation or use the quadratic formula. Since factoring might not be straightforward, let's use the quadratic formula:

x = (-(-2) ± √((-2)² - 4(1)(-4))) / (2(1))

Simplifying:

x = (2 ± √(4 + 16)) / 2

x = (2 ± √20) / 2

x = (2 ± 2√5) / 2

Simplifying further:

x = 1 ± √5

Hence, the solutions to the equation are x = 1 + √5 and x = 1 - √5.