Dạng 3: Tìm số chưa biết trong lũy thừa

Ta có:  $2^x.4=128\Rightarrow 2^x=128:4\Rightarrow 2^x=32\Rightarrow 2^x=2^5\Rightarrow x=5.$

Ta có:  $2^x-26=6\Rightarrow 2^x=6+26\Rightarrow 2^x=32\Rightarrow 2^x=2^5\Rightarrow x=5.$

Ta có:  ${64.4}^x=4^5$

$$\begin{gathered} 4^3{.4}^x=4^5 \\ {4}^{x+3}=4^5 \\ x+3=5 \\ x=5-3 \\ x=2. \end{gathered}$$

Vậy x = 2

Ta có:  ${27.3}^x=243\Rightarrow 3^x=243:27\Rightarrow 3^x=9\Rightarrow 3^x=3^2\Rightarrow x=2.$

Ta có:  ${49.7}^x=2401\Rightarrow 7^x=2401:49\Rightarrow 7^x=49\Rightarrow 7^x=7^2\Rightarrow x=2.$

Ta có:  $3^x=81\Rightarrow 3^x=3^4\Rightarrow x=4.$

Ta có:  $3^4{.3}^x=3^7\Rightarrow 3^x=3^7:3^4\Rightarrow 3^x=3^4\Rightarrow x=4.$

Ta có:  $3^x+25={26.2}^2+{2.3}^0\Rightarrow 3^x=26.1+2.1-25\Rightarrow 3^x=3^1\Rightarrow x=1.$      

Ta có: $3^x.3=243\Rightarrow 3^x=243:3\Rightarrow 3^x=81\Rightarrow 3^x=3^4\Rightarrow x=4.$

Ta có:  $2^x{.16}^2=1024\Rightarrow 2^x=1024:{16}^2\Rightarrow 2^x=1024:256\Rightarrow 2^x=4\Rightarrow 2^x=2^2\Rightarrow x=2.$

Ta có:  ${64.4}^x={16}^8\Rightarrow 4^3{.4}^x={\left(4^2\right)}^8\Rightarrow 4^{x+3}=4^{16}\Rightarrow x+3=16\Rightarrow x=16-3\Rightarrow x=13.$

Ta có:  $2^x=16\Rightarrow 2^x=2^4\Rightarrow x=4.$

$$\begin{gathered} {\left(7x-11\right)}^3=2^5{.5}^2+200 \\ {\left(7x-11\right)}^3=32.25+200 \\ {\left(7x-11\right)}^3=1000 \\ {\left(7x-11\right)}^3={10}^3 \end{gathered}$$

$$\begin{gathered} 7x-11=10 \\ 7x=21 \\ x=3 \end{gathered}$$

Vậy x = 3

Ta có:  $\frac{x-2019}{4}=\frac{1}{x-2019}\Rightarrow {\left(x-2019\right)}^2=4\Rightarrow {\left(x-2019\right)}^2=2^2\Rightarrow x-2019=2\Rightarrow x=2021.$

Ta có:  ${\left(2x-1\right)}^4=16\Rightarrow {\left(2x-1\right)}^4=\pm {\left(2\right)}^4\Rightarrow 2x-1=\pm 2.$ 

TH 1:  $2x-1=2\Rightarrow 2x=3\Rightarrow x=\frac{3}{2}$.

TH 2:  $2x-1=-2\Rightarrow 2x=-1\Rightarrow x=\frac{-1}{2}$.

Vậy  $x=\frac{3}{2}$ hoặc  $x=\frac{-1}{2}.$ 

 ${\left(2x+1\right)}^4={\left(2x+1\right)}^6\Rightarrow {\left(2x+1\right)}^4-{\left(2x+1\right)}^6=0\Rightarrow {\left(2x+1\right)}^4\left[1-{\left(2x-1\right)}^2\right]=0\Rightarrow \left[\begin{array}{l}{\left(2x+1\right)}^4=0\\ 1-{\left(2x+1\right)}^2=0\end{array}\right.$
 $\Rightarrow \left[\begin{array}{l}2x+1=0\\ 2x+1=1\\ 2x+1=-1\end{array}\right.\Rightarrow \left[\begin{array}{l}x=-\mathrm{0,5}\\ x=0\\ x=-1\end{array}\right.$

Vậy  $x=-\mathrm{0,5};x=0;x=-1.$

Ta có:   $\left|\frac{39}{2}-3x^2\right|=\frac{15}{2}\Rightarrow \frac{39}{2}-3x^2=\frac{15}{2}\Rightarrow 3x^2=\frac{39}{2}-\frac{15}{2}\Rightarrow 3x^2=12\Rightarrow x^2=4\Rightarrow x^2={\left(\pm 2\right)}^2\Rightarrow x=\pm 2.$

Ta có:  ${\left(2x+1\right)}^3=125\Rightarrow {\left(2x+1\right)}^3=5^3\Rightarrow 2x+1=5\Rightarrow 2x=5-1\Rightarrow 2x=4\Rightarrow x=4:2\Rightarrow x=2.$

Ta có:  ${\left(3x-1\right)}^{10}={\left(3x-1\right)}^{20}\Rightarrow {\left(3x-1\right)}^{20}-{\left(3x-1\right)}^{10}=0$

 $\begin{array}{l}\Rightarrow {\left(3x-1\right)}^{10}\left[{\left(3x-1\right)}^{10}-1\right]=0\Rightarrow \left[\begin{array}{l}3x-1=0\\ {\left(3x-1\right)}^{10}=1\end{array}\right.\Rightarrow \left[\begin{array}{l}x=\frac{1}{3}\\ 3x-1=1\\ 3x-1=-1\end{array}\right.\Rightarrow \left[\begin{array}{l}x=\frac{1}{3}\\ x=\frac{2}{3}\\ x=0\end{array}\right.\\ \end{array}$

Ta có:    $x{\left(6-x\right)}^{2003}={\left(6-x\right)}^{2003}\Rightarrow x{\left(6-x\right)}^{2003}-{\left(6-x\right)}^{2003}=0$ $\Rightarrow {\left(6-x\right)}^{2003}\left(x-1\right)=0\Rightarrow \left[\begin{array}{l}6-x=0\\ x-1=0\end{array}\right.\Rightarrow \left[\begin{array}{l}x=6\\ x=1\end{array}\right.$

Ta có:    $5^x+5^x{.5}^2=650\Rightarrow 5^x\left(1+25\right)=650\Rightarrow 5^x=25\Rightarrow 5^x=5^2\Rightarrow x=2$

Ta có:  $2^{x+2}-2^x=96\Rightarrow 2^x.4-2^x=96\Rightarrow 2^x.\left(4-1\right)=96\Rightarrow {3.2}^x=96\Rightarrow 2^x=32\Rightarrow 2^x=2^5\Rightarrow x=5$

Ta có:  $2^{x+1}{.3}^y={12}^x\Rightarrow 2^{x+1}{.3}^y=2^2{.3}^x\Rightarrow \left\{\begin{array}{l}x+1=2\\ x=y\end{array}\right.\Rightarrow \left\{\begin{array}{l}x=1\\ y=1\end{array}\right.$

Vậy  $x=y=1.$

Ta có: ${10}^x:5^y={20}^y\Rightarrow {10}^x={20}^y{.5}^y\Rightarrow {10}^x={100}^y\Rightarrow {10}^x={10}^{2y}\Rightarrow x=2y.$