Tìm x:
\(3^x.3^{x+2}=243\)
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`#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) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
(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
3x+3x+1+3x+2=243.39
3x+3x.3+3x.9=9477
3x.(1+3+9)=9477
3x.13=9477
3x=729=36
x=6
3x + 3x+1 + 3x+2 = 243.39
<=> 3x + 3x.3 + 3x.32 = 243.39
<=> 3x( 1 + 3 + 32 ) = 243.39
<=> 3x.13 = 243.39
<=> 3x = 243.3 = 729
<=> 3x = 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
Bài giải
\(3^x\cdot3^{x+2}=243\)
\(3^{x+x+2}=243\)
\(3^{2x+1}=243\)
\(3^{2x+1}=3^5\)
\(\Rightarrow\text{ }2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4\text{ : }2\)
\(x=2\)
3^x.3^x+2=243
Suy ra3^x.3^x.3^2=243
3^x.(3^2.1)=243
3^x.9=243
3^x=243:9
3^x=27
Suy ra x=3