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a. \(\left(2x-3^2\right)=25\Rightarrow2x=25+3^2=25+9=34\)
\(\Rightarrow x=34:2=17\)
Vậy x = 17
b. \(\dfrac{27}{3^x}=3\Rightarrow3^x=27:3=9\Rightarrow3^x=3^2\Rightarrow x=2\)
Vậy x = 2
c. \(9^{x-1}=\dfrac{1}{9}\Rightarrow3^{2x-2}=3^{-2}\)
\(2x-2=-2\Rightarrow2x=-2+2=0\Rightarrow x=0\)
Vậy x = 0
d. \(3^{x+1}+3^{x+3}=810\Rightarrow\left(1+3^2\right)\cdot3^{x+1}=810\)
\(\Rightarrow\left(1+9\right)\cdot3^{x+1}=810\Rightarrow10\cdot3^{x+1}=810\)
\(\Rightarrow3^{x+1}=81\Rightarrow3^{x+1}=3^4\Rightarrow x+1=4\Rightarrow x=3\)
Vậy x = 3
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3. Từ \(\dfrac{x-2}{27}=\dfrac{3}{x-2}\Rightarrow\left(x-2\right)^2=81\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm9\right)^2\\ \Rightarrow\left[{}\begin{matrix}x-2=-9\\x-2=9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=11\end{matrix}\right.\)
Vậy x = -7 hoặc x = 11
4. Từ \(\dfrac{2x+5}{9-2x}=\dfrac{2}{5}\)
\(\Rightarrow5\left(2x+5\right)=2\left(9-2x\right)\\ \Leftrightarrow10x+25=18-4x\\ \Leftrightarrow14x=-7\\ \Rightarrow x=-\dfrac{1}{2}\)
5. Từ \(\dfrac{x-7}{x+8}=\dfrac{x-8}{x+9}\)
\(\Rightarrow\left(x-7\right)\left(x+9\right)=\left(x-8\right)\left(x+8\right)\\ \Leftrightarrow x^2+2x-63=x^2-64\\ \Leftrightarrow2x=-1\\ \Rightarrow x=-\dfrac{1}{2}\)
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a) \(32< 2^x< 128\)
=> \(2^5< 2^x< 2^7\)
=> x = 6
b) \(2^{x-1}+4\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
=> \(2^{x-1}+2^{2+x}=9\cdot2^5\)
=> 9.2x-1 = 9.25
=> 2x-1 = \(\frac{9\cdot2^5}{9}=2^5\)
=> x - 1 = 5 => x = 6
c) \(9\cdot27\le3^x\le243\)
=> \(243\le3^x\le243\)
=> x = 5
d) Giống câu b)
e) \(3^{x-1}+5\cdot3^{x-2}=216\)
=> 8.3x-2 = 216
=> 3x-2 = 27
=> 3x-2 = 33
=> x - 2 = 3 => x = 5
f) 27x-3 = 9x+3
=> 27x-3 = 9x+3
=> (33)x-3 = (32)x+3
=> 33x-9 = 32x + 6
=> không thỏa mãn x vì x là phân số mà theo đề bài là số nguyên
g) x2019 = x => x2019 - x = 0 => x(x2018 - 1) = 0 => x = 0 hoặc x = 1
a)
\(2^5< 2^x< 2^7\)
\(5< x< 7\)
\(x=6\)
b)
\(2^{x-1}+2^2\cdot2^x=9\cdot2^5\)
\(2^{x-1}+2^{2+x}=9\cdot2^5\)
\(2^{x-1}\left(1+2^3\right)=9\cdot2^5\)
\(2^{x-1}\cdot9=9\cdot2^5\)
\(2^{x-1}=2^5\)
\(x-1=5\)
\(x=6\)
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Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)
\(\Rightarrow4^{2x}=8.2^x\)
\(\Rightarrow4^{2x}=2^3.2^x\)
\(\Rightarrow\left(2^2\right)^{2x}=2^{x+3}\)
\(\Rightarrow2^{4x}=2^{x+3}\)
=> 4x = x + 3
=> 3x = 3
=> x = 1
Vậy x = 1.
`9^x : 3^x = 3`
`(9:3)^x = 3`
`3^x = 3`
`3^x=3^1`
`=> x=1`