can anyone solve these 3 exponential functions??PLEASE HELP!!?

Q1

3^(6 - 3x) = 1/27

3^(6 - 3x) = 3^-3 . . . . <= bases are same, so powers are also same

Compare:

6 - 3x = -3

9 = 3x

x = 3

Q2

2^[(x^2) - 3] = 64

2^[(x^2) - 3] = 2^6

Compare:

x^2 - 3 = 6

x^2 = 9

x = ±√9

x = ±3

Q3

9^(2x) × 27^(3 - x) = 1/9

(3^2)^(2x) × (3^3)^(3 - x) = 3^-2

3^(4x) × 3^[3(3 - x)] = 3^-2

3^(4x) × 3^(9 - 3x) = 3^-2

3^(4x + 9 - 3x) = 3^-2

3^(x + 9) = 3^-2

Compare:

x + 9 = -2

x = -11

Done

1. x = 3 -- just take log base 3 of both sides and solve the resulting equation.

2. x = 3. Same as above, except you take log base 2.

3. x = -11. 9 = 3^2 and 27 = 3^3 -- use this to simplify the equation to 3^(x+9) = 1/9. Take log base 3 and solve.

1) 3^(6 - 3x) = 3^(-3)

6 - 3x = -3

-3x = -9

x = 3

2) 2^((x^2)-3) = 64

2^((x^2)-3) = 2^6

x^2 - 3 = 6

x^2 = 9

x = +/- 3

3) 9^(2x) * 27^(3-x)= (1/9)

3^(4x) * 3^(3(3-x) = 3^-2

3^(4x + 3(3 - x)) = -2

4x + 9 - 3x = -2

x = -11

1. x = 3

2. x1 = 3 x2 = -3

3. i GUESS x = -10

i am quite sure for 1 and 2, but 3 i'm not sure, can't be bothered to get a paper :D

use logs. You are solving equations of exponential functions. You are not solving exponential functions.