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a, 2x-1 = 64
=> 2x-1 = 26
=> x - 1 = 6
=> x = 6 + 1
=> x = 7
b, 32x-10 = 81
=> 32x-10 = 34
=> 2x - 10 = 4
=> 2x = 14
=> x = 7
c, 22x+3 = 1024
=> 22x+3 = 210
=> 2x + 3 = 10
=> 2x = 7
=> x = \(\frac{7}{2}\)
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`<=>2^x=1024/162=512/81`
Vì `x\inNN` nên không tồn tại `x` thoả mãn.
Vậy không tồn tại `x` là số tự nhiên thoả mãn đề.
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`@` `\text {Ans}`
`\downarrow`
`2^x = 16`
`=> 2^x = 2^4`
`=> x = 4`
Vậy, `x = 4.`
____
`2^x*16 = 1024`
`=> 2^x =`\(2^{10}\div2^4\)
`=> 2^x = 2^6`
`=> x = 6`
Vậy, `x = 6`
______
`2^x - 26 = 6`
`=> 2^x = 6 + 26`
`=> 2^x = 32`
`=> 2^x = 2^5`
`=> x = 5`
Vậy, `x = 5`
`3^x*3 = 243`
`=> 3^x * 3 = 3^5`
`=> 3^x = 3^5 \div 3`
`=> 3^x = 3^4`
`=> x = 4`
Vậy, `x = 4.`
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(2x-48)^3=8 (10.x)^2=0
(2x)^3-48^3=2^3 10^2-x^2=0^2
2.x-48=2 10-x=0
2.x=48:2 x=10-0
2.x=24 x=10-0
x=24:2 x=10
x=12
Tick cho mìn nha ý cuối hình như bn chép sai đề bài đó
(2x - 48)^3 = 8
Ta có (2x-48)3=23
2x-48=2
2x=50
x=25 Vậy x=25
(10-x)^2=0
suy ra 10-x=0
x=10
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\(a,|2x-2019|=1\)
\(\Leftrightarrow\orbr{\begin{cases}2x-2019=1\\2x-2019=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=2020\\2x=2018\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1010\\x=1009\end{cases}}\)
Vậy ............
\(b,\left(2-x\right)^5=-32\)
\(\Leftrightarrow\left(2-x\right)^5=\left(-2\right)^5\)
\(\Leftrightarrow2-x=-2\)
\(\Leftrightarrow x=4\)
Vậy ..........
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a, 2 x . 2 2 = 32
2 x + 2 = 2 5
x + 2 = 5
x = 3
Vậy x = 3
b, 27 . 3 x = 243
3 3 . 3 x = 3 5
3 3 + x = 3 5
x + 3 = 5
x = 2
Vậy x = 2
c, 2 x . 2 4 = 1024
2 x + 4 = 2 10
x + 4 = 10
x = 6
Vậy x = 6
d, 49 . 7 x = 2401
7 2 . 7 x = 7 4
7 2 + x = 7 4
2 + x = 4
x = 2
Vậy x = 2
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a) \(2^{2x}.2^4=1024\)
\(2^{2x}=1024:2^4\)
\(2^{2x}=1024:16\)
\(2^{2x}=64\)
\(2^{2x}=2^6\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
vay \(x=3\)
b) \(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=2.5.3^{12}+2^3.\left(3^3\right)^4\)
\(2.3^x=2.5.3^{12}+2^3.3^{12}\)
\(2.3^x=2.3^{12}.\left(5+2^2\right)\)
\(2.3^x=2.3^{12}.9\)
\(2.3^x=2.3^{12}.3^2\)
\(2.3^x=2.3^{14}\)
\(\Rightarrow x=14\)
vay \(x=14\)
c) \(5^8.25^x+1=5^{17}\)
\(5^8.\left(5^2\right)^x+1=5^{17}\)
\(5^8.5^{2x}+1=5^{17}\)
\(5^{8+2x}=5^{17}-1\)
e) \(\left(2x-4\right)^5=\left(2x-4\right)^3\)
\(\left(2x-4\right)^5-\left(2x-4\right)^3=0\)
\(\left(2x-4\right)\left[\left(2x-4\right)^2-1\right]=0\)
\(\left(2x-4\right)\left(2x-4-1\right)\left(2x-4+1\right)=0\)
\(\left(2x-4\right)\left(2x-5\right)\left(2x-3\right)=0\)
\(\Rightarrow2x-4=0\)hoac \(\orbr{\begin{cases}2x-5=0\\2x-3=0\end{cases}}\)
\(\Rightarrow2x=4\)hoac \(\orbr{\begin{cases}2x=5\\2x=3\end{cases}}\)
\(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
vay \(x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(2^x-2^y=1024\Rightarrow x>y\)
Do đó \(2^y\left(2^{x-y}-1\right)=2^{10}\)
Lại có \(2^{x-y}-1\) lẻ và là ước 10 nên \(2^{x-y}-1=1\Rightarrow2^y=2^{10}\)
\(\Rightarrow y=10\Rightarrow2^{x-10}=2^1\Rightarrow x=11\)
Vậy \(\left(x;y\right)=\left(11;10\right)\)
\(2^{x+1}\) =1024
=> \(2^{x+1}\) = \(2^{10}\)
=> x+1 = 10
=> x = 10 - 1
=> x = 9
\(2^{x+1}=1024\)
\(2^{x+1}=2^{10}\)
\(\Rightarrow x+1=10\)
\(\Rightarrow x=9\)
k nhé tớ k lại cho hứa đó
hihihihi ^ _ ^ ihihin~