Bài 2:
c) 2 . $5^{x + 1}$ - 1100 = 6 . $5^{2}$
⇔ 2 . $5^{x + 1}$ - 1100 = 6 . 25
⇔ 2 . $5^{x + 1}$ - 1100 = 150
⇔ 2 . $5^{x + 1}$ = 150 + 1100
⇔ 2 . $5^{x + 1}$ = 1250
⇔ $5^{x + 1}$ = 1250 : 2
⇔ $5^{x + 1}$ = 625
⇔ $5^{x + 1}$ = $5^{4}$
⇒ x + 1 = 4
⇔ x = 4 - 1
⇔ x = 3
Vậy x = 3
b) 2 . $2^{2x}$ + $4^{3}$ . $4^{x}$ = 1056
⇔ 2 . $2^{2x}$ + $(2^2)^{3}$ . $(2^2)^{x}$ = 1056
⇔ 2 . $2^{2x}$ + $2^{6}$ . $2^{2x}$ = 1056
⇔ $2^{2x}$ . (2 + $2^{6}$) = 1056
⇔ $2^{2x}$ . (2 + 64) = 1056
⇔ $2^{2x}$ . 66 = 1056
⇔ $2^{2x}$ = 1056 : 66
⇔ $2^{2x}$ = 16
⇔ $2^{2x}$ = $2^{4}$
⇒ 2x = 4
⇔ x = 4 : 2
⇔ x = 2
Vậy x = 2
f) $x^{2}$ = 4
⇔ $x^{2}$ = $2^{2}$ hoặc ⇔ $x^{2}$ = $-2^{2}$
⇒ x = 2 hoặc ⇔ x = -2
Vậy x = 2 hoặc x = -2
h $(2x - 15)^{5}$ = $(2x - 15)^{3}$
⇔ $(2x - 15)^{5}$ - $(2x - 15)^{3}$ = 0
⇔ $(2x - 15)^{3}$ . $(2x - 15)^{2}$ - $(2x - 15)^{3}$ = 0
⇔ $(2x - 15)^{3}$ . {$(2x - 15)^{2}$ - 1} = 0
⇔ $\left \{ {{(2x - 15)^{3} = 0} \atop {(2x - 15)^{2} - 1 = 0 }} \right.$
⇔ x = $\frac{15}{2}$
⇔ x = 8; x = 7
Vậy x = $\frac{15}{2}$; x = 8 hoặc x = 7
