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(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
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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
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(3.x)^2 : 3^3=243
9.x^2 : 27=243
9.x^2=243.27
9.x^2=6561
x^2=6561:9
x^2=729
X=27
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\(8;a,3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
\(=\frac{3^2.\left(3^4\right)^2}{243.3^3}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
\(b,\frac{4.2^5}{2^3.\frac{1}{16}}\)
\(=\frac{2^2.2^5}{2^3.\frac{1}{2^4}}\)
\(=\frac{2^7}{\frac{1}{2}}=2^7.2=2^8\)
a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3}^3\)
\(=3^2.\frac{1}{243}.\left(3^4\right)^2.\frac{1}{27}\)
\(=3^2.\frac{1}{243}.3^8.\frac{1}{27}\)
\(=\frac{3^2.3^8}{243.27}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
b, \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)\)
\(=\left(2^2.2^5\right):\left(8.\frac{1}{16}\right)\)
\(=2^7:\frac{1}{2}\)
\(=2^8\)
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phần b nhé
(X+0,7)3=-27
(X+0,7)3=(-3)3
X+0,7=-3
X=3-0,7=2,3
Vậy...
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3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn
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(3x)2 : 33 = \(\frac{1}{243}\)
<=> 32x : 33 = \(\frac{1}{3^5}\)
<=> 32x - 3 = 3-5
=> 2x - 3 = -5
2x = -5 + 3
2x = -2
x = -2 : 2
x = -1
\(\left(3^x\right)^2:3^3=\frac{1}{243}\Rightarrow3^{2x}:3^3=\frac{1}{3^5}\Rightarrow3^{2x-3}=3^{-5}\Rightarrow2x-3=-5\Rightarrow2x=-2\Rightarrow x=-1\)