Question

Решить уравнение 81^x-8*9^x+15=0

Answer 1

${81}^{x} - 8 \times {9}^{x} + 15 = 0 \\ {9}^{2x} - 8 \times {9}^{x} + 15 = 0 \\ {9}^{x} = t \\ {t}^{2} - 8t + 15 = 0 \\ \frac{d}{4} = 16 - 15 = 1\\ t = 4 + 1 = 5 \\ t = 4 - 1 = 3 \\ {9}^{x} = 3 \\ {3}^{2x} = 3 \\ 2x = 1 \\ x = \frac{1}{2} \\ {9}^{x} = 5 \\ x = log_{9}(5) = log_{ {3}^{2} }(5) = \frac{1}{2} log_{3}(5)$
Ответ :
$x = \frac{1}{2} \\ x = \frac{1}{2} log_{3}(5)$

Answer 2

Сделаем замену 9^x = t, тогда 81^x = t^2.

Получаем:

t^2 - 8t + 15 = 0. Находим корни через теорему Виета. Это t1 = 3 ; t2 = 5 .

Учитывая замену:

9^x = 3 => x = 1/2 .

9^x = 5 => x = log(9) 5 .

Ответ: x = 1/2 ; x = log(9) 5.

Answer 3

To solve the equation 81^x - 8 * 9^x + 15 = 0, we can make a substitution to simplify the equation.

Let's rewrite 81 as (9^2) and 15 as (3 * 5):

(9^2)^x - 8 * 9^x + 3 * 5 = 0

Now, we can make a substitution to simplify the equation:

Let y = 9^x

The equation becomes:

y^2 - 8y + 15 = 0

Now, we can solve this quadratic equation for y:

Using the quadratic formula: y = (-b ± √(b^2 - 4ac)) / 2a

where a = 1, b = -8, and c = 15

y = (8 ± √((-8)^2 - 4 * 1 * 15)) / 2 * 1 y = (8 ± √(64 - 60)) / 2 y = (8 ± √4) / 2 y = (8 ± 2) / 2

Now, we have two possible values for y:

1. y = (8 + 2) / 2 = 10 / 2 = 5
2. y = (8 - 2) / 2 = 6 / 2 = 3

Now, recall that y = 9^x:

1. 5 = 9^x Taking the logarithm base 9 of both sides: x = log_9(5)

2. 3 = 9^x Taking the logarithm base 9 of both sides: x = log_9(3)

So, the two solutions for the equation are:

1. x = log_9(5)
2. x = log_9(3)