2^x+1 x 16^2=1024
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Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
a) 2^x . 16^2 = 1024 b) 64 . 4^x = 16^8 c) 2^x = 16
=> 2^x . 256 = 1024 => 64 . 4^x = (4^2) ^ 8 => 2^x = 2^4
=> 2^x = 1024 : 256 => 4^3 . 4^x = 4^16 => x = 4
=> 2^x = 4 => 4^x = 4^16 : 4^3
=> 2^x = 2^2 => 4^x = 4^13
=> x = 13
=> x = 2
a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)
b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)
c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)
2x.162 = 1024
=> 2x.(24)2 = 210
=> 2x + 4.2 = 10
=> x + 8 = 10
=> x = 2
vậy_
64.4x = 168
=> 43.4x = (42)8
=> 43 + x = 416
=> 3 + x = 16
=> x = 13
vậy_
2x = 16
=> 2x = 24
=> x = 4
vậy_
256 x 6=512 x 3
512 x 3=256 x 6
999 x 2>1000
512 x 5>1024
1024<512 x 5
1024=512 x 2
512=256 x 2
256=128 x 2
128=64 x 2
64=32 x 2
32=16 x 2
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
\(2^x.16^2=1024\)
\(2^x=1024:16^2\)
\(2^x=1024:256\)
\(2^x=4\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
vay \(x=2\)
2x . 162 = 1024
\(\Rightarrow\)2x . 256 = 1024
\(\Rightarrow\)2x = 1024 : 256 = 4
\(\Rightarrow\)2x = 22
\(\Rightarrow\)x = 2
x^20-x=0
x(x^19-1)=0
x= 0
hoặc x ^ 19 =1
x = 0 hoặc x= 1
2^x+5*2^4=2^10
=>2^x+5+4=2^10
=>x+5+4=10
=>x=10-(5+4)
=>x=1
a, 2x.16 = 1024 => 2x = 1024:16 => 2x = 64 => 2x = 26 => x = 6
b, x17 = x
=> x17 - x = 0
=> x(x16-1)=0
=> x = 0 hoặc x16 - 1 = 0
=> x = 0 hoặc x16 = 1
=> x = 0 hoặc x = 1
c, (2x-2)3=64
=> (2x-2)3 = 43
=>2x-2=4
=>2x=6
=>X=3
d,(x-6)2 = (x-6)3
=> (x-6)2-(x-6)3=0
=> (x-6)2-[1-(x-6)] = 0
=> (x-6)2 = 0 hoặc 1 - (x-6) = 0
=> x - 6 = 0 hoặc x - 6 = 1
=> x = 6 hoặc x = 7
e, 3 + 2x-1 = 24-[42-(22-1)]
=> 3 + 2x-1 = 11
=> 2x-1 = 8
=> 2x-1 = 23
=>x-1=3
=>x=4
`2^(x+1) xx 16^2 =1024`
`=> 2^(x+1) xx 2^8 = 2^10`
`=> 2^(x+1) = 2^10 : 2^8`
`=> 2^(x+1) = 2^2`
`=> x+1 =2`
`=> x = 2-1`
`=> x = 1`
Vậy `x=1`
2\(^{x+1}\) \(\times\) 162 = 1024
2\(^{x+1}\) = 1024 : 162
2\(^{x+1}\) = 4
2\(^{x+1}\) = 22
\(x+1\) = 2
\(x=2-1\)
\(x=1\)
Vậy \(x=1\)