Question

9^-x=27 4^3x=8^2x-3 3^2x-3=1/27

Answer 1

Решение на фото........

Answer 2

We have three equations here:

1. 9^(-x) = 27
2. 4^(3x) = 8^(2x-3)
3. 3^(2x-3) = 1/27

Let's solve them one by one:

1. 9^(-x) = 27 Rewriting 27 as a power of 9, we get: 9^(-x) = 9^(3/2) Equating the exponents, we get: -x = 3/2 Therefore, x = -3/2.

2. 4^(3x) = 8^(2x-3) Rewriting 8 as a power of 4, we get: 4^(3x) = (4^2)^(2x-3) Simplifying, we get: 4^(3x) = 4^(4x-6) Taking the logarithm of both sides (with base 4), we get: 3x = 4x-6 Therefore, x = 6.

3. 3^(2x-3) = 1/27 Rewriting 1/27 as a power of 3, we get: 3^(2x-3) = 3^(-3) Equating the exponents, we get: 2x-3 = -3 Therefore, x = 0.

Therefore, the solutions to the system of equations are x = -3/2, 0, and 6.