Question

A.
^2 log (x-2) - ^2 log (2x-5) + ^2 log (x-1) = 1
b.
3^2x+1 + 3 < ^3x+2 3^x

Answer 1

2'log (x-2) - 2'log(2x-5) + 2'log (x+1) = 2'log 2
2'log (x-2)(x+1) / (2x-5) = 2'log 2
(x-2)(x+1)/ (2x-5) = 2
(x-2)(x-1) = 2(2x-5)
x^2 - 3x + 2 = 4x - 10
x^2 - 7x + 12 = 0
(x-4)(x-3) = 0
x = 4 atau x = 3
.
2) 3^(2x+1) + 3  ≤ 3^(x+2) + 3^x
3^2x. 3 + 3  - 3^x.3^2 - 3^x ≤ 0
..
set 3^x = a
3 . a^2 + 3 - 9 a - a ≤ 0
3 a^2 - 10 a + 3 ≤ 0
(3a -1)(a-3) ≤ 0
a = 1/3  atau a = 3
3^x = 1/3---> x = -1
3^x = 3 --> x = 1
HP = (-1, 1)