Question

(2/3)^х-1×(27/8)^2-х=243/32

Answer 1

Запишем в виде
(2/3)^(x-1)×(2/3)^(x-2)=(2/3)^(-5)
(x-1)×(x-2)=-5
x^2-3x+7=0
D=9-28 <0 действительных решений нет

Answer 2

To solve the equation (2/3)^(x-1) * (27/8)^(2-x) = 243/32, we can start by simplifying the exponents and bases on both sides of the equation.

Let's simplify the equation step by step:

1. Rewrite 243/32 as (3/2)^5:

(2/3)^(x-1) * (27/8)^(2-x) = (3/2)^5

2. Now, let's work on the bases:

2/3 = (2^1) / (3^1) 27/8 = (3^3) / (2^3)

So, the equation becomes:

((2^1) / (3^1))^(x-1) * ((3^3) / (2^3))^(2-x) = (3/2)^5

3. Apply the exponent rule (a^m)^n = a^(m*n) to both sides of the equation:

(2^(x-1) * 3^(x-1)) * (3^(6-2x) * 2^(6-2x)) = (3/2)^5

4. Now, simplify the equation further:

2^(x-1) * 3^(x-1) * 3^(6-2x) * 2^(6-2x) = (3/2)^5

5. Use the fact that a^m * a^n = a^(m+n):

2^(x-1 + 6-2x) * 3^(x-1 + 6-2x) = (3/2)^5

6. Simplify the exponents:

2^5 * 3^5 = (3/2)^5

7. Calculate the values on both sides:

32 * 243 = (3/2)^5

8. Calculate (3/2)^5:

(3/2)^5 = 243/32

9. Now, we have:

32 * 243 = 243/32

This equation is satisfied, which means that the original equation is true for all values of x.

So, the solution to the equation (2/3)^(x-1) * (27/8)^(2-x) = 243/32 is x can be any real number, and the equation will hold true.