`#3107.101107`

a)

\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)

Vậy, `x = 4`

b)

\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)

Vậy, `x = 3`

c)

\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)

Vậy, `x = 4.`

a) \(27< 3^x< 243\)

\(\Rightarrow3^3< 3^x< 3^5\)

\(\Rightarrow3< x< 5\)

c) \(3^x+3^{x+2}=810\)

\(\Rightarrow3^x\left(1+3^2\right)=810\)

\(\Rightarrow3^x.10=810\)

\(\Rightarrow3^x=810:10\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

sai đề rồi

\(\Leftrightarrow3^x\cdot3=243\)

hay x=4

(3^x)^2=1/243.3^3

(3^x)^2=1/9

3x=1/3 hoặc 3x=-1/3

Suy ra x=1/9 hoặc x=-1/9

3^x( 1 + 3 + 3²) = 3⁶.13

3^x. 13 = 3⁶.13

=> 3^x= 3⁶

=> x= 6

mk ko hiểu cho lắm

bn ko hiểu chỗ nào???

3^x+3^x+1+3^x+2=243.39

 3^x+3^x.3+3^x.9=9477

     3^x.(1+3+9)=9477

              3^x.13=9477

                   3^x=729=3⁶

                          x=6

3^x + 3^x+1 + 3^x+2 = 243.39

<=> 3^x + 3^x.3 + 3^x.3² = 243.39

<=> 3^x( 1 + 3 + 3² ) = 243.39

<=> 3^x.13 = 243.39

<=> 3^x = 243.3 = 729

<=> 3^x = 3⁶

<=> x = 6

a: =>1/3x-2/5x=5

=>-1/15x=5

=>x=-75
b: =>4x=4

=>x=1

c: =>6*3^x-5*3^x=243

=>3^x=243

=>x=5

\(3^{2x+1}=243\)

\(\Rightarrow3^{2x+1}=3^5\)

\(\Rightarrow2x+1=5\)

\(\Rightarrow2x=5-1\)

\(\Rightarrow2x=4\)

\(\Rightarrow x=4:2\)

\(\Rightarrow x=2\)

\(3^2x+1=243\)

\(9x=243-1\)

\(9x=242\)

\(x=242:9\)

\(\Rightarrow x=\dfrac{242}{9}\)

`a)x^3=343=7^3`

`=>x=7`

Vậy `x=7`

`b)(x-2,5)^4=(x-2,5)^2`

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

`+)(x-2,5)^2=0<=>x=2,5`

`+)(x-2,5)^2=1`

`TH1:x-2,5=1<=>x=3,5`

`th2:x-2,5=-1<=>x=1,5`

Vậy `x=0` hoặc `x=1,5` hoặc `x=3,5

`c)x^8/243=27`

`=>x^8=27.243`

`=>x^8=3^3*3^5=3^8`

`=>x=+-3`

Giải:

a) \(x^3=343\) 

\(\Rightarrow x^3=7^3\) 

\(\Rightarrow x=7\) 

b) \(\left(x-2,5\right)^4=\left(x-2,5\right)^2\) 

Vì có cùng cơ số nhưng lại khác mũ thì \(\left(x-2,5\right)\in\left\{0;1\right\}\) 

TH1:

\(\Rightarrow x-2,5=0\) 

\(\Rightarrow x=2,5\) 

TH2:

\(\Rightarrow\left[{}\begin{matrix}x-2,5=1\\x-2,5=-1\end{matrix}\right.\) 

\(\Rightarrow\left[{}\begin{matrix}x=3,5\\x=1,5\end{matrix}\right.\) 

c) \(\dfrac{x^8}{243}=27\) 

\(\Rightarrow x^8=3^3.3^5\) 

\(\Rightarrow x^8=3^8\) 

\(\Rightarrow x=\pm3\)