Question

solve for x

1. 7=49^x

2. (1/4)^x=16

3.2^x+3 = 9^x+3

4. 3^x+1 = 9^x-2

5. (1/4)^2-x = (1/27)^3+x

Answer 1

1) 7 = 49^x
49 = 7^2, so 49^x = 7^{2x}. Hence 7 = 7^{2x} ⇒ 2x = 1 ⇒ x = 1/2.

2) (1/4)^x = 16
(1/4)^x = 4^{-x} and 16 = 4^2, so -x = 2 ⇒ x = -2.

3) 2^x + 3 = 9^x + 3
Cancel 3: 2^x = 9^x ⇒ (2/9)^x = 1 ⇒ x = 0.

4) 3^x + 1 = 9^x - 2
Let y = 3^x (>0). Then y + 1 = y^2 - 2 ⇒ y^2 - y - 3 = 0.
y = (1 ± √13)/2; only the positive root works:
3^x = (1 + √13)/2 ⇒ x = log_3((1 + √13)/2) ≈ 0.7590.

5) (1/4)^{2-x} = (1/27)^{3+x}
Write as 2^{-4+2x} = 3^{-9-3x}. Take logs:
(−4 + 2x) ln 2 = (−9 − 3x) ln 3
Solve for x:
x = (4 ln 2 − 9 ln 3) / (2 ln 2 + 3 ln 3) ≈ −1.5196.