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Ta có :
\(8< 2^x\le2^9.2^{-5}\)(1)
Xét :
\(2^9.2^{-5}=2^9.\frac{1}{2^5}=2^4\)(2)
Thay (2) vào (1) ta có :
\(\Rightarrow2^3< 2^x\le2^4\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
a) \(8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< x\le2^{9-5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow3< x\le4\Leftrightarrow x=4\)
b) \(27< 81^3:3^x< 243\)
\(\Leftrightarrow3^2< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^2< 3^{12-x}< 3^5\)
\(\Leftrightarrow2< 12-x< 5\)
\(\Leftrightarrow\hept{\begin{cases}x=8\\x=9\end{cases}}\)
a,\(8< 2^x\le2^9.2^{-5}\)
\(2^3< 2^x\le2^4\)
\(\Rightarrow x=4\)
b, \(27< 81^3.3^x< 243\)
\(3^3< 3^{12-x}< 3^5\)
\(\Rightarrow3< 12-x< 5\)
12-x=4
x=8
c,\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^3.\left(\frac{2}{5}\right)^2\)
\(\left(\frac{2}{5}\right)^x>\left(\frac{2}{5}\right)^5\)
\(\Rightarrow x>5\)
x=6;7;8........
\(a)8< 2^x\le2^9.2^{-5}\)
\(\Leftrightarrow2^3< 2^x\le2^9.\dfrac{1}{2^5}\)
\(\Leftrightarrow2^3< 2^x\le2^4\)
\(\Leftrightarrow x=4\)
\(b)27< 81^3:3^x< 243\)
\(\Leftrightarrow3^3< \left(3^4\right)^3:3^x< 3^5\)
\(\Leftrightarrow3^3< 3^{12}:3^x< 3^5\)
\(\Leftrightarrow3^3< 3^{12-x}< 3^5\)
\(\Leftrightarrow12-x=4\)
\(\Leftrightarrow x=8\)
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}\le1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
<=> \(\frac{41}{36}\le\frac{x}{36}\le\frac{35}{24}\)
<=> \(\frac{82}{72}\le\frac{2x}{72}\le\frac{105}{72}\)
<=> \(82\le2x\le105\)
<=> \(41\le x\le52,5\)
Do \(x\in N\)nên \(x=\left\{x\in N|41\le x\le52,5\right\}\)
a)\(32^{-n}\cdot16^n=2048\)
\(\left(2^5\right)^{-n}\cdot\left(2^4\right)^n\)=2048
\(2^{-5n}\cdot2^{4n}\)=\(2^{11}\)
\(2^{-5n+4n}=2^{11}\)
\(2^{-x}=2^{11}\)
\(\Rightarrow x=-11\)
b)\(2^{-1}\cdot2^n+4\cdot2^n=9\cdot2^5\)
\(\frac{1}{2}\cdot2^n+4\cdot2^n=288\)
\(2^n\left(\frac{1}{2}+4\right)=288\)
\(2^n\cdot\frac{9}{2}=288\)
\(2^n=288:\frac{9}{2}\)
\(2^n=64\)
\(2^n=2^6\)
\(\Rightarrow n=6\)
a) 32-n . 16n = 2048
\(\frac{1}{32n}\) . 16n = 2048
\(\frac{1}{2^n.16^n}\) . 16n = 2048
\(\frac{1}{2^n}\) = 2048
2-n = 2048
2-n = 211
\(\Rightarrow\) -n = 11
\(\Rightarrow\) n = -11
Vậy n = -11
8<2x<29.2-5
23 < 2x< 24
=> x thuộc rỗng ( viết bằng kí hiệu )