Question

Log_5(x^2-8)=log_5(2-9x)

Answer 1

Находим ОДЗ 1)x^2-8>0 x>2sqrt2 или x<-sqrt2 2)2-9x>0 x<2/9 3)Основание логарифмов равны,значит приравниваем подлогарифмичиские выражения,имеем x^2-8=2-9x x^2+9x-10=0 По теореме Виета корни квадратного уравнения равны x=-10 x=1(по Одз x=1 не подходит) Ответ х=-10

Answer 2

To solve this equation, we can start by using the logarithmic property that states:

log_a(b) = log_a(c) if and only if b = c, for any positive base a.

Applying this property to the given equation, we get:

log_5(x^2-8) = log_5(2-9x) x^2 - 8 = 2 - 9x (since log_5 is an increasing function) x^2 + 9x - 10 = 0

Now we can solve this quadratic equation using the quadratic formula:

x = (-b ± sqrt(b^2 - 4ac)) / 2a

where a = 1, b = 9, and c = -10.

x = (-9 ± sqrt(9^2 - 4(1)(-10))) / 2(1) x = (-9 ± sqrt(121)) / 2 x = (-9 ± 11) / 2

So the solutions are:

x = (-9 + 11) / 2 = 1 x = (-9 - 11) / 2 = -10

However, we need to check if these solutions satisfy the original equation. We can see that x = -10 does not satisfy the equation, because the argument of the logarithm on the left-hand side becomes negative. Therefore, the only solution is x = 1.