Question

Log2(x)+log2(x-1)=log2(6)

Answer 1

log_2(x) + log_2( x - 1 ) = log_2(6)

ОДЗ: х > 0 ;  х - 1 > 0 ⇒ х > 1  

Воспользуемся свойством логарифмов: log_a(b) + log_a(c) = log_a(b•c)

log_2( x( x - 1 ) ) = log_2(6)

В силу монотонного возрастания функции у = log_a(x) , где а > 1, следует:

х•( х - 1 ) = 6

х² - х - 6 = 0

х₁ = 3

х₂ = - 2

С учётом ОДЗ  ⇒  х = 3

ОТВЕТ: х = 3

Answer 2

To solve the equation log2(x) + log2(x-1) = log2(6), we can use the properties of logarithms, specifically the product rule of logarithms, which states that:

logb(x) + logb(y) = logb(xy)

Using this property, we can simplify the left-hand side of the equation as follows:

log2(x) + log2(x-1) = log2(x(x-1))

Now we can substitute this expression back into the original equation, giving:

log2(x(x-1)) = log2(6)

Since the logarithm function is one-to-one, this equation is equivalent to:

x(x-1) = 6

Expanding the left-hand side and rearranging terms, we get:

x^2 - x - 6 = 0

This is a quadratic equation that can be factored as:

(x-3)(x+2) = 0

Therefore, the solutions to the equation log2(x) + log2(x-1) = log2(6) are x = 3 and x = -2. However, since the logarithm function is defined only for positive values of its argument, we reject the solution x = -2. Therefore, the only solution to the equation is x = 3.