1.

a)9.$27^{n}$=$3^{5}$

$3^{2}$.$27^{n}$=$3^{5}$

$3^{3n}$ =$3^{5}$:$3^{2}$

$3^{3n}$ =$3^{3}$

3n=3

n=1

b)($2^{3}$:4).$2^{n}$=4

($2^{3}$:$2^{2}$).$2^{n}$=$2^{2}$

2. $2^{n}$=$2^{2}$

$2^{n}$=2

n=2

c)$3^{-2}$.$3^{4}$.$3^{n}$=$3^{7}$ 

$3^{2}$.$3^{n}$ =$3^{7}$

$3^{n}$=$3^{5}$

n=5

d)$2^{-1}$.$2^{n}$+4. $2^{n}$=9.$2^{5}$ 

$2^{n-1}$+$2^{2}$.$2^{n}$=9.$2^{5}$

$2^{n-1}$.(1+$2^{3}$)=9.$2^{5}$

$2^{n-1}$.9=9.$2^{5}$

$2^{n-1}$=$2^{5}$ 

n-1=5

n=6

2.

a)125.5≥$5^{n}$≥5.25

$5^{3}$.5≥$5^{n}$ ≥5.$5^{2}$ 

$5^{4}$ ≥$5^{n}$ ≥$5^{3}$ 

=>4≥n≥3

=>n∈{4;3}

b)($n^{54}$)$^{2}$ =n

$n^{108}$ =n

$n^{108}$-n=0

n.($n^{107}$ -1)=0

+)n=0

+)$n^{107}$-1=0

$n^{107}$=1

n=1

c)243≥$3^{n}$ ≥9.27

$3^{5}$≥$3^{n}$ ≥$3^{2}$.$3^{3}$ 

$3^{5}$≥$3^{n}$ ≥$3^{5}$

5≥n≥5

n=5

Câu d mik ko làm nhé.

3.

a)$2^{300}$=($2^{3}$)$^{100}$=$8^{100}$ 

$3^{200}$=($3^{2}$)$^{100}$ =$9^{100}$ 

mà $8^{100}$ <$9^{100}$ 

nên $2^{300}$<$3^{200}$

Mik có việc bận tối mik làm tiếp cho nhé.

Bài 1:

a) 9.$27^{n}$ = $3^{5}$ 

<=> 9.$27^{n}$ = 243

<=> $27^{n}$ =. 27

Vậy n = 1

b) ($2^{3}$ : 4).$2^{n}$ =4

<=> (8:4). $2^{n}$ = 4

<=> 2. $2^{n}$ =4

<=> $2^{n}$ = 2

Vậy n =1

c) $3^{-2}$. $3^{4}$. $3^{n}$ =$3^{7}$ 

<=> $\frac{1}{9}$ . 81 . $3^{n}$ = 2187

<=> 9. $3^{n}$ = 2187

<=> $3^{n}$=243

Vậy n = 5.

d) $2^{-1}$ .$2^{n}$+4. $2^{n}$= 9. $2^{5}$ 

<=> $\frac{1}{2}$. $2^{n}$ + 4.$2^{n}$ =288

<=> $2^{n}$ . ( $\frac{1}{2}$ +4)=288

<=> $2^{n}$ . 4,5 = 288

<=>$2^{n}$ = 64

Vậy n= 6

Bài 3:

a) $2^{300}$và $3^{200}$ 

Ta có:

$2^{300}$ = ($2^{3}$)^100 = $8^{100}$ 

$3^{200}$ = ($3^{2}$)^100 = $9^{100}$

Vì $8^{100}$ < $9^{100}$ => $2^{300}$ < $3^{200}$.

b) $99^{20}$ và  $9999^{10}$ 

Ta có: $99^{20}$ = $99^{2.10}$ = $9081^{10}$ 

Vì $9081^{10}$ < $9999^{10}$ =>$99^{20}$ < $9999^{10}$ 

c) $3^{500}$ và $7^{300}$

Ta có: $3^{500}$ = ($3^{5}$) ^100 = $243^{100}$

$7^{300}$ = ($7^{3}$)^100 = $343^{100}$

Vì 243 < 343 => $3^{500}$ < $7^{300}$