a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)

\(27^n:3^n=\left(27:3\right)^n=9\)

\(9^n=9\rightarrow n=1\)

\(\left(\frac{25}{5}\right)^n=5^n=5^1\)

\(\rightarrow n=1\)

\(\frac{81}{\left(-3\right)^n}=-243=\left(-3\right)^5\)

\(\rightarrow\left(-3\right)^n=81:\left(-3\right)^5=\frac{-1}{3}=\left(-3\right)^{-1}\)

\(\)

a) 27^n : 3^n = 9

     (27 : 3)^n  = 9

         9^n        = 9

=> n = 1

b) 25/5^n = 5

       5^n     = 25 : 5

       5^n     = 5

=> n = 1

c) 81/(-3)^n = -243

         (-3)^n   = -243 : 81

         (-3)^n   = -3

=> n = 1

d) 1/2 . 2^n + 4 . 2^n = 9 . 2^5

       2^n . (1/2 + 4)     = 9 . 32

       2^n . 9/2              = 288

       2^n                       = 288 : 9/2

       2^n                       = 64

       2^n                       = 2^6

=> n = 6

thank you for doing your homework well !

\(a)27^n\div3^n=9\)

\(\Rightarrow\left(27\div3\right)^n=9\)

\(\Rightarrow9^n=9\Rightarrow n=1\)

\(b)\frac{25}{5^n}=5\Rightarrow25=5^{n+1}\Rightarrow5^2=5^{n+1}\Rightarrow2=n+1\Rightarrow n=1\)

\(c)\frac{81}{\left(-3\right)^n}=-243\Rightarrow\frac{81}{\left(-3\right)^n}=\left(-3\right)^5\)

\(\Rightarrow81=\left(-3\right)^{n+5}\Rightarrow\left(-3\right)^4=\left(-3\right)^{n+5}\)

\(\Rightarrow4=n+5\Rightarrow n=-1\)

\(d)\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=9.2^5\Rightarrow2^n=9.2^5.\frac{2}{9}\Rightarrow2^n=\frac{9.2^5.2}{9}=2^6\)

\(\Rightarrow n=6\)

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

a) \(3^3\)

b)\(2^8\)

c) \(2^7\)

d) \(3^1\)

a) 9.3³.\(\dfrac{1}{81}\) .3² = 3². 3³.\(\dfrac{1}{3^4}\) . 3² = 3³

b) 4. 2⁵: \(\) (2³.\(\dfrac{1}{16}\))= 2². 2⁵: 2³. \(\dfrac{1}{2^4}\) = 2⁷: \(\dfrac{1}{2}\) = 2⁷. 2= 2⁸

c) 3². 2⁵. \(\left(\dfrac{2}{3}\right)^2\) = 3². 2⁵. \(\dfrac{2^2}{3^2}\) = 2⁵. 2² = 2⁷

d) \(\left(\dfrac{1}{3}\right)^2\) .\(\dfrac{1}{3}\) . 9² = \(\dfrac{1}{9}.\dfrac{1}{3}\). 9² = \(\dfrac{9}{3}\) = 3¹

a không bít !

b chịu

c quên rồi

d . . .

a)27ⁿ:3ⁿ=9

   (27:3)ⁿ=9

   9ⁿ=9¹

   n=1

Vậy n=1

b)\(\left(\frac{25}{5}\right)^n=5\)

    \(5^n=5^1\)

    n=1

Vạy n=1

c)\(\left(-\frac{81}{3}\right)^n=-243\)

   \(\left(-27\right)^n=\left(-3\right)^5\)

   \(\left[\left(-3\right)^3\right]^n=\left(-3\right)^5\)

   \(\left(-3\right)^{3n}=\left(-3\right)^5\)

    \(3n=5\)

   \(n=\frac{5}{3}\)

Vậy \(n=\frac{5}{3}\)

d)\(\frac{1}{2}.2^n+4.2^n=9.5^n\)

   \(2^n.\left(\frac{1}{2}+4\right)=9.5^n\)

   \(2^n.\frac{9}{2}=3^2.5^n\)

27^n:3^n=9

3^3n:3^n=3^2

3^(3n-n)=3^2

suy ra 3n-n=2

n(3-1)=2

n.2=2

Vậy n=2:2

n=1

còn câu sau cậu tự tính

\(n=1\)

a,Ta có \(\left(3^3\right)^n:3^n=9\Leftrightarrow3^{3n}:3^n=3^2\Leftrightarrow3n-n=2\Leftrightarrow n=1\)

b,TA có \(\dfrac{5^2}{5^n}=5^1\Leftrightarrow2-n=1\Leftrightarrow n=1\)

Các câu sau để bn tự làm

a) 27ⁿ : 3ⁿ = 9

\(\Leftrightarrow\) (27 : 3)ⁿ = 9

\(\Leftrightarrow\) 9ⁿ = 9

\(\Leftrightarrow\) n = 1

b) \(\dfrac{25}{5^n}=5\)

\(\Leftrightarrow\dfrac{5^2}{5^n}=5\)

\(\Leftrightarrow5^n.5=5^2\)

\(\Leftrightarrow5^{n+1}=5^2\)

\(\Leftrightarrow n+1=2\)

n = 2 - 1

n = 1

c) \(\dfrac{81}{\left(-3\right)^n}=-243\)

\(\Leftrightarrow\dfrac{\left(-3\right)^4}{\left(-3\right)^n}=\left(-3\right)^5\)

\(\Leftrightarrow\left(-3\right)^n.\left(-3\right)^5=\left(-3\right)^4\)

\(\Leftrightarrow\left(-3\right)^{n+5}=\left(-3\right)^4\)

\(\Leftrightarrow n+5=4\)

n = 4 - 5

n = -1