a)1/9.27^n=3^n

             3^n=3^n

         =>n={0;1;2;3;...}     

a) n= 2;3;5;7;...(n là số nguyên)

 a)1/9.27^n=3^n

 3^n=3^n

=>n={0;1;2;3...}

 Tích nha ^_^ !!!

a)

1/9 . 3^4.3ⁿ=3⁷

=>3^-2.3^4.3ⁿ=3⁷

=>3^-2+4+n=3⁷

=>-2+4+n=7

=>n=7-(-2)-4

=>n=5

a Ta có

1/9.3^4.3^n=3^7

=> 1/3^2.3^4.3^n=3^7

=> 3^2.3^n=3^7

=>3^2+n=3^7

=> 2+n=7

=> n=5( tick nhé)

1/9 . 27^n=3^n

=1/3^2.3^3^n=3^n

=3^2.3^3=3^n

=3^5

=>n=5

a) 32 < 2ⁿ > 128

<=> 2⁵ < 2ⁿ > 2⁷

<=> n = 8 ; 9 ; 10...

b) 2 . 16 < 2ⁿ > 4

<=> 2¹ . 2⁴ < 2ⁿ > 4

<=> 2⁵ < 2ⁿ > 4

<=> n = 5 ; 6 ; 7 ;...

c) ( 2² : 4 ) . 2ⁿ = 4

<=> 1 . 2ⁿ = 4

<=> 2ⁿ = 4

<=> 2ⁿ = 2²

<=> n = 2

Bài 1:

a. $2^{29}< 5^{29}< 5^{39}$

$\Rightarrow A< B$

b.

$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$

$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$

$=(1+3)(3+3^3+3^5+...+3^{2009})$

$=4(3+3^3+3^5+...+3^{2009})\vdots 4$

Mặt khác:

$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$

$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$

$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$

Bài 1:
c.

$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$

$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$

$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$

$\Rightarrow A=\frac{3^{101}+1}{4}$