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

`#3107`

b)

`2.3^x = 162`

`\Rightarrow 3^x = 162 \div 2`

`\Rightarrow 3^x = 81`

`\Rightarrow 3^x = 3^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

c)

`(2x - 15)^5 = (2 - 15)^3`

\(\Rightarrow \)`(2x - 15)^5 - (2x - 15)^3 = 0`

\(\Rightarrow \)`(2x - 15)^3 . [ (2x - 15)^2 - 1] = 0`

\(\Rightarrow\left[{}\begin{matrix}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=15\\\left(2x-15\right)^2=\left(\pm1\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x-15=1\\2x-15=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\2x=16\\2x=-14\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=-7\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-7;8;\dfrac{15}{2}\right\}.\)

`d)`

\(3^{x+2}-5.3^x=?\) Bạn ghi tiếp đề nhé!

`e)`

\(7\cdot4^{x-1}+4^{x-1}=23?\)

\(4^{x-1}\cdot\left(7+1\right)=23\\ \Rightarrow4^{x-1}\cdot8=23\\ \Rightarrow4^{x-1}=\dfrac{23}{8}\)

Bạn xem lại đề!

`f)`

\(2\cdot2^{2x}+4^3\cdot4^x=1056\)

\(\Rightarrow2\cdot2^{2x}+\left(2^2\right)^3\cdot\left(2^2\right)^x=1056\\ \Rightarrow2\cdot2^{2x}+2^6\cdot2^{2x}=1056\\ \Rightarrow2^{2x}\cdot\left(2+2^6\right)=1056\\ \Rightarrow2^{2x}\cdot66=1056\\ \Rightarrow2^{2x}=1056\div66\\ \Rightarrow2^{2x}=16\\ \Rightarrow2^{2x}=2^4\\ \Rightarrow2x=4\\ \Rightarrow x=2\)

Vậy, `x = 2`

_____

\(10 -{[(x \div 3+17) \div 10+3.2^4] \div 10}=5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=10-5\)

\(\Rightarrow\left[\left(x\div3+17\right)\div10+48\right]\div10=5\)

\(\Rightarrow\left(x\div3+17\right)\div10+48=50\)

\(\Rightarrow\left(x\div3+17\right)\div10=2\)

\(\Rightarrow x\div3+17=20\)

\(\Rightarrow x\div3=3\\ \Rightarrow x=9\)

Vậy, `x = 9.`

`@` `\text {Ans}`

`\downarrow`

`2^x * 4 = 128`

`=> 2^x = 128 * 4`

`=> 2^x = 512`

`=> 2^x = 2^9`

`=> x = 9`

Vậy, `x = 9`

`x^15 = x`

`=> x^15 - x = 0`

`=> x(x^14 - 1) = 0`

`=>` TH1: `x = 0`

`TH2: x^14 - 1 = 0`

`=> x^14 = 1`

`=> x = 1`

Vậy, `x \in {0; 1}`

`(2x+1)^3 = 125`

`=> (2x+1)^3 = 5^3`

`=> 2x + 1 = 5`

`=> 2x = 5 - 1`

`=> 2x =4`

`=> x = 4 \div 2`

`=> x = 2`

Vậy,` x = 2.`

`(x - 5)^4 = (x-5)^6`

`=> (x-5)^4 - (x-5)^6 = 0`

`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`

`=> - (x-6)(x-5)^4(x-4) = 0`

`TH1: (x - 5)^4 = 0`

`=> x - 5 = 0`

`=> x = 0 +5`

`=> x = 5`

`TH2: x - 6=0`

`=> x=6`

`TH3: x-4=0`

`=> x = 4`

Vậy, `x \in {4; 5; 6}`

a: =>2^x=32

=>x=5

b: =>x^15-x=0

=>x(x^14-1)=0

=>x=0; x=1;x=-1

c: =>2x+1=5

=>2x=4

=>x=2

d: =>(x-5)^4[(x-5)^2-1]=0

=>(x-5)(x-4)(x-6)=0

=>x=5;x=4;x=6

2^x . 4 = 128

2^x      = 128 : 4

2^x      = 32

2^x      = 2 . 2 . 2 . 2 . 2

2^x        = 2⁵

x       = 5

(2x + 1)³ = 125

(2x + 1)³ = 5 . 5 . 5

(2x + 1)³ = 5³

2x + 1    = 5

2x         = 5 - 1 

2x         = 4

x          = 4 : 2

x          = 2

x¹⁵ = x

x = 1

(x - 5)⁴ = (x - 5)⁶

x = 6

a,x=5

b,x=2

c,x=6

tk mk nhé

125=5 mũ 3; suy ra 2x+1=5;2x=4 suy ra x=2

x=1 vì 1 mũ n luôn =1

chịu nhé

a) \(4x+4=16\)

\(4x=12\)

\(x=3\)

b) \(34\left(2x-6\right)=0\)

\(2x=6\)

\(x=3\)

c) \(15:x=5\)

\(x=15:5=3\)

d) \(20-\left(x+14\right)=5\)

\(x+14=20-5=15\)

\(x=15-14=1\)

a) \(4x+4=16\)

\(\Rightarrow4x=16-4\)

\(\Rightarrow4x=12\)

\(\Rightarrow x=\dfrac{12}{4}\)

\(\Rightarrow x=3\)

b) \(34\cdot\left(2x-6\right)=0\)

\(\Rightarrow2x-6=\dfrac{0}{36}\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=\dfrac{6}{2}\)

\(\Rightarrow x=3\)

c) \(15:x=5\)

\(\Rightarrow x=15:5\)

\(\Rightarrow x=3\)

d) \(20-\left(x+14\right)=5\)

\(\Rightarrow x+14=20-5\)

\(\Rightarrow x+14=15\)

\(\Rightarrow x=15-14\)

\(\Rightarrow x=1\)

a)\(2x+3\left(x+4\right)-14=8\)

\(2x+3x+12-14=8\)

\(2x+3x-2=8\)

\(5x-2=8\)

\(5x=8+2\)

\(5x=10\)

\(x=10:5\)

\(x=2\)

b)\(x+2x+3x=4x+16\)

\(x\left(1+2+3\right)=4x+16\)

\(6x-4x=16\)

\(2x=16\)

\(x=16:2\)

\(x=8\)

c)\(2\left(x+1\right)+3\left(x+5\right)=0\)

\(2x+2+3x+15=0\)

\(5x+17=0\)

\(5x=0-17\)

\(5x=-17\)

\(x=-17:5\)

\(x=-\frac{17}{5}\)

mình sẽ k cho bạn nào có câu trả lời ddaayd đủ và đúng nhất 

câu 2:

(2x+1)³=125

(2x+1)³=5³

=>2x+1=5

=>2x=5-1

=>2x=4

=>x=2

hc tốt!

bạn nào giải được tất cả các câu thì mình mới k cho nhé 

thời gian trong sáng thứ 5 khoảng 6 giờ cũng được nhé