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16x < 1284
= (24)x < (27)4
= 24x < 228
=> 4x < 28
= 4x < 4 . 7
=> x < 7
Vậy: x thuộc {0; 1; ... ; 6}
ta co 128^4=16^7
Để 16^x<128^4 thi x<7
vi x la so tu nhien nen x thuoc tap hop cac so 0,1,2,3,4,5,6
![](https://rs.olm.vn/images/avt/0.png?1311)
16x<1284
<=> (24)x < (27)4
<=> 2 4x< 228
=> 4x < 28 => x < 7
Đáp số: x < 7
![](https://rs.olm.vn/images/avt/0.png?1311)
a/16^x < 128^4
=> (4^2)^x < 128^4
=> 4^2x < 128^4
=> (2^2)^2x < 128^4
=> 2^4x < (2^7)^4
=> 2^4x < 2^28
=> 4x < 28
=> x < 7
vậy x = {0,1,2,3,4,5,6}
![](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)
=>16x<1284
=>(2x)4<(27)4
=>2x<27
=>x<7
Vì x thuộc N => x thuộc {0;1;2;3;4;5;6}
Ta có : \(16^x< 128^4\left(x\in N\right)\)
\(\Rightarrow x\in\left\{0;1;2;3;4;5;6\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
16x<1284
=> 16x<1632
Vì n\(\in\)N => x<32 => x\(\in\){0;1;2;3;...;31}
x\(\in\)N* 16x < 1284
16x=(24)x=24x
1284=(27)4=24.7
Vì 24x < 24.7
Nên x<7
Vậy x\(\in\){0;1;2;3;4;5;6}
2x*4=128
\(\Rightarrow2^x\cdot2^2=128\)
\(\Rightarrow2^{x+2}=128\)
\(\Rightarrow2^{x+2}=2^7\)
\(\Rightarrow x+2=7\)
\(\Rightarrow x=5\)
\(2^x\cdot4=128\)
\(2^x=\frac{128}{4}=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)