Dạng 4: Một số bài tập nâng cao về lũy thừa

Ta có:    $3^{2n}={\left(3^2\right)}^n=9^n$

                    $2^{3n}={\left(2^3\right)}^n=8^n$

Vì  $9^n>8^n$ nên  $3^{2n}>2^{3n}$

Ta có:  $S=1+2+2^2+2^3+\mathrm{...}+2^9$

Mà  $2^{10}-1<2^{10}={4.2}^8<{5.2}^8$ 

Vậy  $S<{5.2}^8$.

Ta có:  $A=\frac{{10}^{15}+1}{{10}^{16}+1}$ $\Rightarrow 10A=10.\left(\frac{{\text{10}}^{\text{15}}+1}{{10}^{16}+1}\right)$ =  $\frac{{10}^{16}+10}{{10}^{16}+1}$ =  $\frac{{10}^{16}+1+9}{{10}^{16}+1}=1+\frac{9}{{10}^{16}+1}$.       

           $B=\frac{{10}^{16}+1}{{10}^{17}+1}$  $\Rightarrow 10B=10.\left(\frac{{\text{10}}^{\text{16}}+1}{{10}^{17}+1}\right)$ =  $\frac{{10}^{17}+10}{{10}^{17}+1}$=  $\frac{{10}^{17}+1+9}{{10}^{17}+1}=1+\frac{9}{{10}^{17}+1}$.

Vì   ${10}^{16}+1<{10}^{17}+1$ nên   $\frac{9}{{10}^{16}+1}>\frac{9}{{10}^{17}+1}$ $\Rightarrow 1+\frac{9}{{10}^{16}+1}>1+\frac{9}{{10}^{17}+1}$

 $10A>10B$ hay  $A>B$ 

Ta có:   $C=\frac{2^{2008}-3}{2^{2007}-1}$  $\Rightarrow \frac{1}{2}C=\frac{1}{2}\left(\frac{2^{2008}-3}{2^{2007}-1}\right)=\frac{2^{2008}-3}{2^{2008}-2}=\frac{2^{2008}-2-1}{2^{2008}-2}=1-\frac{1}{2^{2008}-2}$ .

             $D=\frac{2^{2007}-3}{2^{2006}-1}$  $\Rightarrow \frac{1}{2}D=\frac{1}{2}\left(\frac{2^{2007}-3}{2^{2006}-1}\right)=\frac{2^{2007}-3}{2^{2007}-2}=\frac{2^{2007}-2-1}{2^{2007}-2}=1-\frac{1}{2^{2007}-2}$ 

Vì   $2^{2008}–2 > 2^{2007}–2$ nên   $\frac{1}{2^{2008}-2}<\frac{1}{2^{2007}-2}$

           $\Rightarrow 1-\frac{1}{2^{2008}-2}>1-\frac{1}{2^{2007}-2}$

          $\frac{1}{2}C>\frac{1}{2}D$ hay   $C>D.$

Vậy  $C>D.$

Ta giải từng bất đẳng thức  $3^{64}< n^{48}$ và  $n^{48}<5^{72}$.

Ta có:  $n^{48}>3^{64}\Rightarrow {\left(n^3\right)}^{16} > {\left(3^4\right)}^{16} \Rightarrow {\left(n^3\right)}^{16} > {81}^{16}\Rightarrow n^{3 }> 81$

 $\Rightarrow n>4$ (với  $n\in \mathbb{Z}$)                                                    (1).

Mặt khác   $n^{48}<5^{72} \Rightarrow {\left(n^2\right)}^{24} <{\left(5^3\right)}^{24}\Rightarrow {\left(n^2\right)}^{24} <{125}^{24}\Rightarrow n^2 <125$

 $\Rightarrow -11\le n\le 11$  (với  $n\in \mathbb{Z}$)                                         (2).

Từ (1) và (2)$\Rightarrow 4<n\le 11$ .

Vậy $n$  nhận các giá trị nguyên là:  $5;6;7;8;9;10;11.$

Ta có: ${16}^x<{128}^4$$\Rightarrow$ ${\left(2^4\right)}^x<{\left(2^7\right)}^4$$\Rightarrow$ $2^{4x}<2^{28}\Rightarrow 4x<28\Rightarrow x<7$

           $\Rightarrow x\in \left\{\mathrm{0,1,2,3,4,5,6}\right\}$.