Log_5(2x-1)+log_5(4-x)=1

Предыдущий
вопрос Следующий
вопрос

To solve this logarithmic equation, we can use the properties of logarithms to combine the two logarithmic terms on the left side of the equation.

Applying this property to the given equation, we get:
log_5[(2x-1)(4-x)] = 1

Rewrite the equation in exponential form to eliminate the logarithm:
5^1 = (2x-1)(4-x)

Solve the equation:
5 = 8x - 2 - x^2

Rearrange the equation into a quadratic form:
x^2 - 8x + 7 = 0

Factorize the quadratic equation:
(x - 7)(x - 1) = 0

Set each factor to zero:
x - 7 = 0 or x - 1 = 0

Solve for x:
x = 7 or x = 1

Therefore, the solutions to the equation log_5(2x-1) + log_5(4-x) = 1 are x = 7 and x = 1.