a) \(2^x+2^{x+3}=144\)

\(2^x\left(1+2^3\right)=144\)

\(2^x.9=144\)

\(2^x=144:9=16\)

=> \(2^x=2^4\Rightarrow x=4\)

b) \(2^{x-1}+5.2^{x-2}=224\)

\(2^{x-2}\left(2+5\right)=224\)

\(2^{x-2}.7=224\Rightarrow2^{x-2}=32\Rightarrow2^{x-2}=2^5\)

=> x - 2 = 5 => x = 7

a ) 2^x + 2^{x + 3} = 144
=> 2^x . ( 1 + 2³ ) = 144

=> 2^x . 9  = 144

=> 2^x = 16

=> 2^x = 2⁴

=> x = 4
Vậy x = 4

b ) 2^{x - 1}  + 5. 2^{x - 2} = 224

=> 2^{x - 2} . ( 2 + 5 ) = 224

=> 2^{x - 2} . 7 = 224

=> 2^{x - 2} =  32

=> 2^{x - 2} = 2⁵

=> x - 2 = 5

=> x = 7

Vậy x = 7

8^x=2¹¹¹

=>(2³)^x=2¹¹¹

=>2^3x=2¹¹¹

=>3x=111

=>x=37

2^x+5.2^x=48

=>2^x.6=48

=>2^x=8=2³

=>x=3

Câu cuối ko làm đc nha

a. 8^x = 2¹¹¹

    2^3x = 2¹¹¹

    3x  = 111

      x  = 111:3

     x  = 37

b. 2^x + 5.2^x = 48

    2^x(5 + 1 )= 48

    2^x.6        = 48

    2^x           = 48 : 6

    2^x           = 8

    2^x            = 2³

      x            = 3

c. 2^x + 3.2^x-1 = 40

   2^x(3:2 +1 )     = 40

   2^x . \(\frac{5}{2}\)      = 40

    2^x                  = 40 : \(\frac{5}{2}\)

    2^x                  = 16

    2^x                   = 2⁴

      x                   = 4

+) \(8^x=2^{111}\)

\(\Rightarrow\left(2^3\right)^x=2^{111}\)

\(\Rightarrow2^{3x}=2^{111}\)

\(\Rightarrow3x=111\)

\(\Rightarrow x=111:3\)

\(\Rightarrow x=37\)

Vậy x = 37

+) \(2^x+5\times2^x=48\)

\(\Rightarrow2^x\times\left(1+5\right)=48\)

\(\Rightarrow2^x\times6=48\)

\(\Rightarrow2^x=48:6\)

\(\Rightarrow2^x=8\)

\(\Rightarrow2^x=2^3\)

\(\Rightarrow x=3\)

Vậy x = 3

+) \(2^x+3\times2^{x-1}=40\)

\(\Rightarrow2^x+3\times2^x:2=40\)

\(\Rightarrow2^x+3\times2^x\times\frac{1}{2}=40\)

\(\Rightarrow2^x+2^x\times\left(3\times\frac{1}{2}\right)=40\)

\(\Rightarrow2^x+2^x\times\frac{3}{2}=40\)

\(\Rightarrow2^x\times\left(1+\frac{3}{2}\right)=40\)

\(\Rightarrow2^x\times\frac{5}{2}=40\)

\(\Rightarrow2^x=40:\frac{5}{2}\)

\(\Rightarrow2^x=16\)

\(\Rightarrow2^x=2^4\)

\(\Rightarrow x=4\)

Vậy x = 4

_Chúc bạn học tốt_

\(2^x+5\cdot2^{x+2}=168\)

=>\(2^x+20\cdot2^x=168\)

=>\(21\cdot2^x=168\)

=>\(2^x=8=2^3\)

=>x=3

\(2^x+5.2^{x+2}=168\\ \Leftrightarrow2^x.\left(1+5.2^2\right)=168\\ \Leftrightarrow2^x.21=168\\ \Leftrightarrow2^x=\dfrac{168}{21}=8=2^3\\ Vây:x=3\)

2x+5.2x+2=168⇔2x.(1+5.22)=168⇔2x.21=168⇔2x=16821=8=23

Vây:x=3

Sai đề thì phải bạn ơi!

Chắc vậy

Mn jup mk vs

3_² ((x + 4)-5)² = 5x2²

9 (( x + 4 ) - 5)² = 20

????

Không cho điều kiện của x sao làm đc bạn

\(3^{2\left(x+4\right)}-5^2=5.2^2\)

\(3^{2\left(x+4\right)}-25=5.4\)

\(3^2.3^{x+4}=45\)

\(3^{x+4}=9\)

\(3^{x+4}=3^2\)

\(\Rightarrow x+4=2\)

\(x=-2\)

Vay ....

Bài 1:

a) \(4^{x+2}+4^x=68\)

\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)

\(\Rightarrow4^x\cdot17=68\)

\(\Rightarrow4^x=\dfrac{68}{17}\)

\(\Rightarrow4^x=4\)

\(\Rightarrow4^x=4^1\)

\(\Rightarrow x=1\)

b) \(5\cdot2^{x+4}-3\cdot2^x=308\)

\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)

\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)

\(\Rightarrow2^x\cdot77=308\)

\(\Rightarrow2^x=\dfrac{308}{77}\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

c) \(4\cdot3^{x+1}+7\cdot3^x=513\)

\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)

\(\Rightarrow3^x\cdot19=513\)

\(\Rightarrow3^x=\dfrac{513}{19}\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

d) \(5^{x+4}-5^x=3120\)

\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)

\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)

\(\Rightarrow5^x\cdot624=3120\)

\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)

\(\Rightarrow5^x=5\)

\(\Rightarrow5^x=5^1\)

\(\Rightarrow x=1\)

f) \(3\cdot4^{2x+1}-16^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)

\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)

\(\Rightarrow4^{2x}\cdot11=2816\)

\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)

\(\Rightarrow4^{2x}=256\)

\(\Rightarrow\left(2^2\right)^{2x}=2^8\)

\(\Rightarrow2^{4x}=2^8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

Bài 2:

\(2^x+124=5^y\)

\(\Rightarrow5^y-2^x=124\)

\(\Rightarrow5^y-2^x=125-1\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)

Vậy: .... 

a: =>x-8=27

hay x=35