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\(2^x.16^2=1024\)
\(2^x=1024:16^2\)
\(2^x=1024:256\)
\(2^x=4\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
vay \(x=2\)
2x . 162 = 1024
\(\Rightarrow\)2x . 256 = 1024
\(\Rightarrow\)2x = 1024 : 256 = 4
\(\Rightarrow\)2x = 22
\(\Rightarrow\)x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^x\cdot16^2=1024\)
\(\Leftrightarrow2^x=4\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
1024 : ( 17 x 25 + 15 x 25 )
= 1024 : [( 17 + 15 ) x 25 ]
= 1024 : ( 32 x 25 )
= 210 : 210
= 1
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64 . 4x = 168
<=> 43. 4x = 416
=> 3 + x = 16
<=> x = 13
Vậy x = 13
2x.162 = 1024
<=> 2x. 28 = 210
=> x + 8 = 10
<=> x = 2
Vậy x = 2
b: Ta có: \(2^x\cdot16^2=1024\)
\(\Leftrightarrow2^x\cdot2^8=2^{10}\)
\(\Leftrightarrow x+8=10\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
Có: \(\dfrac{2^x}{2^{2012}}=1024\)
\(\Rightarrow\dfrac{2^x}{2^{2012}}=2^{10}\Rightarrow2^x=2^{10}\cdot2^{2012}=2^{2022}\)
\(\Rightarrow x=2022\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2.4^x - 1 = 128
=> 4^x - 1 = 64
=> x - 1 = 3
=> x = 4
b, 17.4^2 + 83.16
= 17.16 + 83.16
= 16(17 + 83)
= 16.100
= 1600
c, a^5 . 10^10.a^15
= a^20.10^10
d, 3{65 - [5(4 + 2.3) - 15] : 7}
= 3{65 - [5(4 + 6) - 15] : 7}
= 3{65 - [5.10 - 15] : 7}
= 3{65 - [50 - 15] : 7}
= 3{65 - 35 : 7}
= 3{65 - 5}
= 3.60
= 180
![](https://rs.olm.vn/images/avt/0.png?1311)
+) \(x^2=121\)
\(\Rightarrow\orbr{\begin{cases}x^2=11^2\\x^2=\left(-11\right)^2\end{cases}}\Rightarrow\orbr{\begin{cases}x=11\\x=-11\end{cases}}\)
Vậy x = 11 hoặc x = -11
+) \(2^{x+3}=1024\)
\(\Rightarrow2^{x+3}=2^{10}\)
\(\Rightarrow x+3=10\)
\(\Rightarrow x=10-3\)
\(\Rightarrow x=7\)
Vậy x = 7
+) \(5^{x+1}=625\)
\(\Rightarrow5^{x+1}=5^4\)
\(\Rightarrow x+1=4\)
\(\Rightarrow x=4-1\)
\(\Rightarrow x=3\)
Vậy x = 3
_Chúc bạn học tốt_
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
2xx15 = 1024
2x = 1024 : 15
2x = ???
Hình như đề sai rồi bn