tìm x biết
2^2 . 3^3x – 27^x = 9^5 b,2^x : 2 = 256 c,(x^2)^4 . 16 = 2^20
tìm x biết 2^2 . 3^3x – 27^x = 9^5 b,2^x : 2 = 256 c,(x^2)^4 . 16 = 2^20
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Philomena
Đáp án:
Giải thích các bước giải:
a) `2^2 . 3^(3x) – 27^x = 9^5`
`=> 4 . 3^(3x) – 3^(3x) = (3^2)^5`
`=> 3^(3x) . (4 – 1) = 3^10`
`=> 3^(3x) . 3 = 3^10`
`=> 3x + 1 = 10`
`=> 3x = 9`
`=> x = 3`
b) `2^x : 2 = 256`
`=> 2^(x – 1) = 2^8`
`=> x – 1 = 8`
`=> x = 8 + 1 = 9`
c) `(x^2)^4 . 16 = 2^20`
`=> x^8 . 2^4 = 2^20`
`=> x^8 = 2^20 : 2^4`
`=> x^8 = 2^16`
`=> x^8 = (2^2)^8`
`=> x^8 = 4^8`
`=> x = 4`
Hoc tốt. Nocopy
Sigridomena
$a$) $2^2 . 3^{3x} – 27^x = 9^5$
$⇔ 4 . 3^{3x} – (3^3)^x = (3^2)^5$
$⇔ 4 . 3^{3x} – 3^{3x} = 3^{10}$
$⇔ 3^{3x} . (4 – 1) = 3^{10}$
$⇔ 3^{3x} . 3 = 3^{10}$
$⇔ 3^{3x+1} = 3^{10}$
$⇔ 3x+1=10$
$⇔ 3x = 9$
$⇔ x = 3$.
Vậy $x=3$
$b$) $2^x : 2 = 256$
$⇔ 2^x : 2 = 2^8$
$⇔ 2^x = 2^8 . 2$
$⇔ 2^x = 2^9$
$⇔ x = 9$
Vậy $x=9$.
$c$) $(x^2)^4 . 16 = 2^{20}$
$⇔ x^8 . 2^4 = 2^{20}$
$⇔ x^8 = 2^{20} : 2^4$
$⇔ x^8 = 2^{16}$
$⇔ x^8 = (2^2)^8$
$⇔ x = 2^2$
$⇔ x = 4$
Vậy $x=4$.