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Bài 1:
2\(x\) = 4
2\(^x\) = 22
\(x=2\)
Vậy \(x=2\)
Bài 2:
2\(^x\) = 8
2\(^x\) = 23
\(x=3\)
Vậy \(x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x = 43 : 25
2x = (22)3 : 25
2x = 26 : 25
2x = 2
=> x = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9\)
\(A=5\cdot\left(2^2\right)^{15}\cdot\left(3^2\right)^9-2^2\cdot3^{20}\cdot\left(2^3\right)^9\)
\(A=5\cdot2^{30}\cdot3^{18}-2^2\cdot3^{20}\cdot2^{27}\)
\(A=5\cdot2^{30}\cdot3^{18}-2^{29}\cdot3^{20}\)
\(A=2^{29}\cdot3^{18}\cdot\left(5\cdot2^1\cdot1-1\cdot3^2\right)\)
\(A=2^{29}\cdot3^{18}\cdot\left(5-9\right)\)
\(A=-2^2\cdot2^{29}\cdot3^{18}\)
\(A=-2^{31}\cdot3^{18}\)
_______________
\(B=5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6\)
\(B=5\cdot2^9\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot\left(3^3\right)^6\)
\(B=5\cdot2^{28}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}\)
\(B=2^{28}\cdot3^{18}\cdot\left(5\cdot1\cdot3-7\cdot2\cdot1\right)\)
\(B=2^{28}\cdot3^{18}\cdot\left(15-14\right)\)
\(B=2^{28}\cdot3^{18}\)
Ta có: \(A:B\)
\(=\left(-2^{31}\cdot3^{18}\right):\left(2^{28}\cdot3^{18}\right)\)
\(=\left(-2^{31}:2^{28}\right)\cdot\left(3^{18}:3^{18}\right)\)
\(=-2^3\cdot1\)
\(=-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^8.2^{20}=2^{16}.2^{20}=2^{36}\)
\(9^{12}.27^5.81^4=3^{24}.3^{15}.3^{16}=3^{55}\)
mk chỉnh đề
\(64^3.4^5.16^2=4^9.4^5.4^4=4^{18}\)
\(25^{20}.125^4=5^{40}.5^{12}=5^{52}\)
\(x^7.x^4.x^3=x^{14}\)
a) 4^x = 8^12
(2^2)^x = (2^3)^12
2^2x = 2^36
=> 2x = 36
=> x = 36 : 2 = 18
a) 4^x=8^12
4^x=(2^3)^12
4^x=2^36
4^x=4^18
=> x=18 hoặc -18
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