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\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)
\(27^n:3^n=\left(27:3\right)^n=9\)
\(9^n=9\rightarrow n=1\)
\(\left(\frac{25}{5}\right)^n=5^n=5^1\)
\(\rightarrow n=1\)
\(\frac{81}{\left(-3\right)^n}=-243=\left(-3\right)^5\)
\(\rightarrow\left(-3\right)^n=81:\left(-3\right)^5=\frac{-1}{3}=\left(-3\right)^{-1}\)
\(\)
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
\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)
a) 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)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)\(\frac{1}{2}.2^n+4.2^n=9.5^n\)
\(2^n.\left(\frac{1}{2}+4\right)=9.5^n\)
\(2^n.\frac{9}{2}=3^2.5^n\)
Mấy dấu chấm là dấu nhân đó 
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
\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)
\(\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
\(\Rightarrow n=3\)
\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)
\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
\(\Rightarrow n=4\)
a, \(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)
Vì \(\dfrac{1}{3}\ne-1,\dfrac{1}{3}\ne0;\dfrac{1}{3}\ne1\) nên \(n=3\)
Vậy........
b, \(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)
Vì \(\dfrac{3}{5}\ne-1,\dfrac{3}{5}\ne0;\dfrac{3}{5}\ne1\) nên \(n=4\)
Vậy..........
Chúc bạn học tốt!!!
a: =>9^n=9
=>n=1
b: =>5^n=5
=>n=1
c: \(\Leftrightarrow\left(-27\right)^n=-243\)
=>\(\left(-3\right)^{3n}=\left(-3\right)^5\)
=>3n=5
=>n=5/3
d: =>2^n*9/2=9*2^5
=>2^n=9*2^5:9/2=2^5*2=2^6
=>n=6