im a little stuck on these 3 functions and cant seem to solve them
1. 3^(6-3x)= (1/27)
2. 2^((x^2)-3)=64
if it looks a little complicated its just 2 raised to x squared minus 3 equals 64
3. 9^(2x) * 27^(3-x)= (1/9)
the star is supposed to be a multiplication symbol
Eric
Favorite Answer
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
PhasedAvalon
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.
Captain Matticus, LandPiratesInc
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
KravaMocna
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
Anonymous
use logs. You are solving equations of exponential functions. You are not solving exponential functions.