\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)

\(\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow n=3\)

\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)

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

\(\Rightarrow n=4\)

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!!!

\(3S=241+81+27+9+...+\dfrac{1}{9}+\dfrac{1}{27}\)

\(2S=3S-S=241-\dfrac{1}{81}=\dfrac{241x81-1}{81}\)

\(\Rightarrow S=\dfrac{241x81-1}{2x81}\)

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

a. \(\dfrac{32}{2^n}=2\)

\(\Leftrightarrow2^n.2=32\)

\(\Rightarrow2^{n+1}=2^5\)

\(\Rightarrow n+1=5\)

\(\Rightarrow n=4\)

Vậy...

b. \(16^n:2^n=8\)

\(\Rightarrow\left(2^4\right)^n:2^n=2^3\)

\(\Rightarrow4n-n=3\)

\(\Rightarrow3n=3\)

\(\Rightarrow n=1\)

Vậy...

c. \(\dfrac{\left(-3\right)^n}{81}=\left(-27\right)\)

\(\Leftrightarrow\left(-3\right)^n=81.\left(-27\right)\)

\(\Rightarrow\left(-3\right)^n=\left(-2187\right)\)

\(\Rightarrow\left(-3\right)^n=\left(-3\right)^7\)

\(\Rightarrow n=7\)

Vậy...

a, \(\dfrac{32}{2^n}\) =2 =>2⁵=2ⁿ⁺¹=>5=n+1=>n=4

b, 2⁴ⁿ:2ⁿ=2³=>2³ⁿ=2³=>3n=3=>n=1

c, \(\dfrac{\left(-3\right)^n}{81}\)=(-27)=>(-3)ⁿ=3⁴ . (-3)³=>n=7

a)x ∈ ∅

b) x=3

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

Tham khảo link:    https://olm.vn/hoi-dap/detail/55111422944.html

`(x+1/3)+(x+1/9)+(x+1/27)+(x+1/81)=56/81`

`x+x+x+x+1/3+1/9+1/27=56/81-1/81`

`4x+13/27=55/81`

`4x=55/81-13/27`

`4x=55/81-52/81`

`4x=16/81`

`x=4/108`

Vậy `x=4/108`

\(A=\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\\ \Rightarrow3A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\\ \Rightarrow3A-A=1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}-\dfrac{1}{3}-\dfrac{1}{9}-\dfrac{1}{27}-\dfrac{1}{81}-\dfrac{1}{243}-\dfrac{1}{729}\\ \Rightarrow2A=1-\dfrac{1}{729}\\ \Rightarrow2A=\dfrac{728}{729}\\ \Rightarrow A=\dfrac{364}{729}\)

Đặt A = \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\)

3A = 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\)

3A - A = ( 3 + 1 + \(\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}\) ) - ( \(1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}+\dfrac{1}{243}+\dfrac{1}{729}\) )

2A = 3 - \(\dfrac{1}{729}=\dfrac{728}{729}\)

A = \(\dfrac{728}{729}:2=\dfrac{364}{729}\)

1093/729

Úi

2A = 3 - \(\dfrac{1}{729}=\dfrac{2186}{729}\) 

A = \(\dfrac{2186}{729}:2=\dfrac{1093}{729}\)

\(x+\dfrac{40}{27}=2\)
\(x=\dfrac{14}{27}\)

\(x+1+\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{1}{27}+\dfrac{1}{81}=2\)

\(\Leftrightarrow x+\dfrac{121}{81}=2\)

hay \(x=\dfrac{41}{81}\)

a, \(\dfrac{16}{2^n}=2\)

\(2^n=\dfrac{16}{2}\)

\(2^n=8\)

\(2^n=2^3\)

=> n = 3

b, \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\left(-3\right)^n=-27\cdot81\)

\(\left(-3\right)^n=\left(-3\right)^3\cdot3^4\)

\(\left(-3\right)^n=\left(-3\right)^7\)

=> n = 7

c, \(8^n:2^n=4\)

\(2^{3n}:2^n=2^2\)

\(2^{2n}=2^2\)

=> 2n = 2

n = 2:2

n = 1

a )

\(\dfrac{16}{2^n}=2\) \(\Leftrightarrow16:x=2\)

\(\Rightarrow x=8\)

\(2^n=8\Rightarrow n=3\)

b )

\(\dfrac{\left(-3\right)^n}{81}=-27\) \(\Leftrightarrow x=-27.81\)

\(\Rightarrow x=-2187\)

\(\left(-3\right)^n=-2187\Rightarrow n=7\)

c )

\(8^n:2^n=4\Leftrightarrow4^n=4\)

\(\Rightarrow n=1\)

(2^2:4)*2^n=4

3^2*3^4*3^n=3^7