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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2^x.4=128\)
\(\Leftrightarrow2^x=128:4\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(b,2^x=4.128\)
\(\Leftrightarrow2^x=2^2.2^7\)
\(\Leftrightarrow2^x=2^9\)
\(\Rightarrow x=9\)
\(c,3^x=3^3.3^5\)
\(\Leftrightarrow3^x=3^8\)
\(\Rightarrow x=8\)
\(d,2^x.\left(2^2\right)^3=\left(2^3\right)^2\)
\(\Leftrightarrow2^x.2^6=2^6\)
\(\Leftrightarrow2^x=2^6:2^6\)
\(\Leftrightarrow2^x=1\)
\(\Rightarrow x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
3:
a: 3^x*3=243
=>3^x=81
=>x=4
b; 2^x*16^2=1024
=>2^x=4
=>x=2
c: 64*4^x=16^8
=>4^x=4^16/4^3=4^13
=>x=13
d: 2^x=16
=>2^x=2^4
=>x=4
![](https://rs.olm.vn/images/avt/0.png?1311)
(\(x\) - 5)2 = 16
(\(x-5\))2 = 42
\(\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
Vậy \(x\) \(\in\) { 1; 9 }
B. 4\(x\) + 33 = 72
4\(^x\) + 33 = 49
4\(^x\) = 49 - 33
4\(^x\) = 16
4\(^x\) = 16
4\(^x\) = 42
\(x\) = 2
c. 2\(^x\).4 = 128
2\(^x\) = 128 : 4
2\(^x\) = 32
2\(^x\) = 25
\(x\) = 5
D, 5\(^x\) . 3 - 75 = 0
5\(^x\).3 = 75
5\(^x\) = 75: 3
5\(^x\) = 25
5\(^x\) = 52
\(x\) = 2
A) \(\left(x-5\right)^2=16=4^2\Rightarrow\left[{}\begin{matrix}x-5=4\\x-5=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)
B) \(4^x+33=7^2\Rightarrow4^x=49-33\Rightarrow4^x=16=4^2\Rightarrow x=2\)
C) \(2^x.4=128\Rightarrow2^x=32=2^5\Rightarrow x=5\)
D) \(5^x.3-75=0\Rightarrow5^x.3=75\Rightarrow5^x=25=5^2\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
128 - ( x + 13 ) = 87
x + 13 = 128 - 87
x + 13 = 41
x = 41 - 13
x = 28
2 . x - 13 = 29
2 . x = 29 - 13
2 .x = 16
x = 16 : 2
x = 8
128 - 3 . x = 92
3 . x = 128 - 92
3 . x = 36
x = 36 : 3
x = 12
546 : ( x + 17 ) = 13
x + 17 = 546 : 13
x + 17 = 42
x = 42 - 17
x = 25
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 2x : 4 = 128
=> 2x = 512
=> 2x = 29
=> x = 9
b) x15 = x
=> x15 = x15
=> x = 1 hoặc 0
c) ( 2x + 1 )3 = 125
=> ( 2x + 1 )3 = 53
=> 2x + 1 = 5
=> 2x = 4
=> x = 2
a) 2x : 4 = 128
=> 2x = 128 . 4
=> 2x = 512
=> 2x = 29
=> x = 9
b) x15 = x
=> x15 = x15
=> x = 1 hoặc 0
c) ( 2x + 1 )3 = 125
=> ( 2x + 1 )3 = 53
=> 2x + 1 = 5
=> 2x = 5 - 1
=> 2x = 4
=> x = 4 : 2
=> x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
128-3(x-5)=24+1010
128 - 3 × ( x - 5 ) = 16 + 1 ( vì tất cả các lũy thừa có số mũ là 0 đều = 1 )
128 - 3 × ( x - 5 ) = 17
3 × ( x - 5 ) = 128 - 17
3 × ( x - 5 ) = 111
( x - 5 ) = 111 ÷ 3
x - 5 = 37
x = 37 + 5
x = 42
Chúc bạn học tốt!