\(a)8< 2^x\le2^9.2^{-5}\)

\(\Leftrightarrow2^3< 2^x\le2^9.\dfrac{1}{2^5}\)

\(\Leftrightarrow2^3< 2^x\le2^4\)

\(\Leftrightarrow x=4\)

\(b)27< 81^3:3^x< 243\)

\(\Leftrightarrow3^3< \left(3^4\right)^3:3^x< 3^5\)

\(\Leftrightarrow3^3< 3^{12}:3^x< 3^5\)

\(\Leftrightarrow3^3< 3^{12-x}< 3^5\)

\(\Leftrightarrow12-x=4\)

\(\Leftrightarrow x=8\)

\(P/s:\)\(Bạn\) \(tự\) \(kết\) \(luận\) \(nha\)

a) ko có x ???

v) 27<81³:3^x>243

3³<(3⁴)³:3^x>3⁵

3³<3¹²:3^x>3⁵

3³<3^12-x>3⁵

=> 3^12-x=3⁴

<=> 12-x=4 

=>x=8

sửa lại cái dấu > thành < nha  bạn

"x" ở đâu vậy bạn?

Sửa đề: \(9\cdot27< =\left(\dfrac{1}{3}\right)^x< =27\cdot243\)

=>3⁵<=3^-x<=3⁸

=>5<=-x<=8

=>\(x\in\left\{-6;-7;-8;-5\right\}\)

mk chắc chắn 100% là n >1

b)\(27< 3^n< 243\)

\(3^3< 3^n< 3^5\)

\(\Rightarrow3< n< 5\)

\(\Rightarrow n\in\left\{4\right\}\)

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........

tìm x, biết:

(5x+1)^2=36/49

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) \(27< 81^3:3^x< 243\)

\(\Leftrightarrow3^2< \left(3^4\right)^3:3^x< 3^5\)

\(\Leftrightarrow3^2< 3^{12}:3^x< 3^5\)

\(\Leftrightarrow3^2< 3^{12-x}< 3^5\)

\(\Leftrightarrow2< 12-x< 5\)

\(\Leftrightarrow\hept{\begin{cases}x=8\\x=9\end{cases}}\)

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

minh bik lam bai b thui a

25^x=5^2.x:5^1.x+y=5^3

suy ra 2.x : 1.x+y=3

a) \(\dfrac{1}{8}.16^n=2^n\)

=> \(2^{4.n-3}\)=\(2^n\)

=> \(4.n-3=n\)

=> \(n=1\)

b) \(27< 3^n< 243\)

=> \(3^3< 3^n< 3^5\)

=> \(n=4\)