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1: =>3^x=81
=>x=4
2: =>2^x=8
=>x=3
3: =>x^3=2^3
=>x=2
4: =>x^20-x=0
=>x(x^19-1)=0
=>x=0 hoặc x=1
5: =>2^x=32
=>x=5
6: =>(2x+1)^3=9^3
=>2x+1=9
=>2x=8
=>x=4
7: =>x^3=115
=>\(x=\sqrt[3]{115}\)
8: =>(2x-15)^5-(2x-15)^3=0
=>(2x-15)^3*[(2x-15)^2-1]=0
=>2x-15=0 hoặc (2x-15)^2-1=0
=>2x-15=0 hoặc 2x-15=1 hoặc 2x-15=-1
=>x=15/2 hoặc x=8 hoặc x=7
1. Tìm số tự nhiên x biết:
1) \(3^x.3=243\)
\(3^x=243:3\)
\(3^x=81\)
\(3^x=3^4\)
\(\Rightarrow x=4\)
_____
2) \(7.2^x=56\)
\(2^x=56:7\)
\(2^x=8\)
\(2^x=2^3\)
\(\Rightarrow x=3\)
_____
3) \(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=3\)
_____
4) \(x^{20}=x\)
\(x^{20}-x=0\)
\(x\left(x^{19}-1\right)=0\)
\(\Rightarrow x=0\) hoặc \(x=1\)
5) \(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
\(2^x=2^5\)
\(\Rightarrow x=5\)
_____
6) \(\left(2x+1\right)^3=9.81\)
\(\left(2x+1\right)^3=729=9^3\)
\(\rightarrow2x+1=9\)
\(2x=9-1\)
\(2x=8\)
\(x=8:2\)
\(\Rightarrow x=4\)
_____
7) \(x^6:x^3=125\)
\(x^3=125\)
\(x^3=5^3\)
\(\Rightarrow x=5\)
_____
8) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=7\\x=8\end{matrix}\right.\)
_____
9) \(3^{x+2}-5.3^x=36\)
\(3^x.\left(3^2-5\right)=36\)
\(3^x.\left(9-5\right)=36\)
\(3^x.4=36\)
\(3^x=36:4\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
_____
10) \(7.4^{x-1}+4^{x+1}=23\)
\(\rightarrow7.4^{x-1}+4^{x-1}.4^2=23\)
\(4^{x-1}.\left(7+4^2\right)=23\)
\(4^{x-1}.\left(7+16\right)=23\)
\(4^{x-1}.23=23\)
\(4^{x-1}=23:23\)
\(4^{x-1}=1\)
\(4^{x-1}=4^1\)
\(\rightarrow x-1=0\)
\(x=0+1\)
\(\Rightarrow x=1\)
Chúc bạn học tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, x - 3 : 2 = 5 14 : 5 12
=> x - 3 : 2 = 5 2
=> x - 3 : 2 = 25
=> x – 3 = 25
=> x = 53
b, 30 : x - 7 = 15 19 : 15 18
=> 30 : x - 7 = 15
=> x – 7 = 2
=> x = 9
c, x 70 = x
=> x 70 - x = 0
=> x ( x 69 - 1 ) = 0
=>
d, 2 x + 1 3 = 9 . 81
=> 2 x + 1 3 = 9 3
=> 2x + 1 = 9
=> x = 4
e, 5 x + 5 x + 2 = 650
=> 5 x 1 + 5 2 = 650
=> 5 x . 26 = 650
=> 5 x = 25
=> x = 2
f, 4 x - 1 2 = 25 . 9
=> 4 x - 1 2 = 5 2 . 3 2
=> 4 x - 1 2 = 15 2
=> 4x – 1 = 15
=> x = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) bạn xem thử có số nào mũ 3 lên bằng 981 không nếu ra đọc đi mình giải
b) \(5^x+5^{x+2}=650\)
\(5^x+5^x+5^2=650\)
\(5^x\left(1+25\right)=650\)
\(5^x.26=650\)
\(5^x=25\)
\(5^x=5^2\)
Vậy x = 2
c) \(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
Vậy x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(2x+1\right)^3=9.81\)
\(\Rightarrow\left(2x+1\right)^3=729\)
\(\Rightarrow\left(2x+1\right)^3=9^3\)
\(\Rightarrow2x+1=9\)
\(\Rightarrow2x=9-1\)
\(\Rightarrow2x=8\)
\(\Rightarrow x=8:2\)
\(\Rightarrow x=4\)
b)\(5^x+5^{x+2}=650\)
\(\Rightarrow5^x+5^x.5^2=650\)
\(\Rightarrow5^x\left(1+25\right)=650\)
\(\Rightarrow5^x.26=650\)
\(\Rightarrow5^x=25\)
\(\Rightarrow x=2\)
c)\(2.3^{2x-1}=54\)
\(\Rightarrow2.3^{2x-1}=54\)
\(\Rightarrow2.\left(3^2\right)^x:3^1=54\)
\(\Rightarrow2.9^x:3=54\)
\(\Rightarrow9^x=81\)
\(\Rightarrow x=2\)
a) 5x + 5x + 2 = 650
=> 5x + 5x.52 = 650
=> 5x + 5x.25 = 650
=> 5x.(1 + 25) = 650
=> 5x.26 = 650
=> 5x = 25
=> 5x = 52
=> x = 2
Vậy x = 2
b) (2x + 1)3 = 9.81
=> (2x + 1)3 = 792
=> (2x + 1)3 = 93
=> 2x + 1 = 9
=> 2x = 8
=> x = 4
Vậy x = 4
c) 2x + 2x + 3 = 144
=> 2x + 2x . 23 = 144
=> 2x + 2x.8 = 144
=> 2x .(1 + 8) = 144
=> 2x.9 = 144
=> 2x = 16
=> 2x = 24
=> x = 4
d) x15 = x2
=> x15 - x2 = 0
=> x2.(x13 - 1) = 0
=> \(\orbr{\begin{cases}x^2=0\\x^{13}-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=0^2\\x^{13}=1^{13}\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
Vậy x = 0 hoặc x = 1
e) 32x + 2 = 9x + 3
=> 32(x + 1) = 9x + 3
=> (32)x + 1 = 9x + 3
=> 9x + 1 = 9x + 3
=> 9x + 3 - 9x + 1 = 0
=> 9x . 93 - 9x.9 = 0
=> 9x 729 - 9x . 9 = 0
=> 9x.720 = 0
=> 9x = 0
=> x \(\in\varnothing\)