Đề bị lỗi. Bạn coi lại đề.

1) \(2^x=4^3 \Leftrightarrow2^x=2^6\Leftrightarrow x=6\)

2) \(2^x=4^6\Leftrightarrow2^x=2^{12}\Leftrightarrow x=12\)

3) \(3^x=9^{10}\Leftrightarrow3^x=3^{20}\Leftrightarrow x=20\)

1) $2^x=4^3\iff 2^x=2^6\iff x=6$

2) $2^x=4^6\iff 2^x=2^{12}\iff x=12$

3) $3^x=9^{10}\iff 3^x=3^{20}\iff x=20$

( 3x - 2⁴ ) . 7³ = 2 . 7⁴

3x - 16 = 2 . 7⁴ : 7³ = 14

3x = 14 + 16 = 30

x = 10

\(\left(3\times x-2^4\right)\times7^3=2\times7^4\)

 \(\left(3\times x-2^4\right)\div2=7^4\div7^3\) 

  \(\left(3\times x-16\right)\div2=7\)

           \(3\times x-16=7\times2\) 

           \(3\times x-16=14\) 

                     \(3\times x=14+16\)                      

                      \(3\times x=30\)

                           \(x=30\div3\) 

                           \(x=10\)

  Bài 1:

2^\(x\) = 4

2\(^x\) = 2²

 \(x=2\)

Vậy \(x=2\)

Bài 2:

2\(^x\) = 8

2\(^x\) = 2³

\(x=3\)

Vậy \(x=3\)

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b: Ta có: \(2^{x+3}+2^x=144\)

\(\Leftrightarrow2^x\cdot9=144\)

\(\Leftrightarrow2^x=16\)

hay x=4

a) (x ^ 54)^2 = x                                         

         x^108  = x

Để: x^108  = x 

=> x=0 hoặc x=1

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