\(2^{x+1}+2^{x+3}=40\)

\(\Leftrightarrow2^x.2+2^x.2^3=40\)

\(\Leftrightarrow2^x.\left(2+8\right)=40\)

\(\Leftrightarrow2^x.10=40\)

\(\Leftrightarrow2^x=40:10\)

\(\Leftrightarrow2^x=4\)

\(\Leftrightarrow x=2\)

Vậy x=2

\(2x\left(2x-1\right)-\left(2x+5\right)^2=0\)

=>\(4x^2-2x-4x^2-20x-25=0\)

=>-22x-25=0

=>22x+25=0

=>22x=-25

=>\(x=-\dfrac{25}{22}\)

a) Biến đổi về dạng (x - 3)(x + 2) = 0. Tìm được x  ∈ { - 2 ; 3 }

b) Thu gọn về dạng -2x + 3 = 0. Tìm được x = 3 2

`@` `\text {Ans}`

`\downarrow`

`2^x = 16`

`=> 2^x = 2^4`

`=> x = 4`

Vậy, `x = 4.`

____

`2^x*16 = 1024`

`=> 2^x =`\(2^{10}\div2^4\)

`=> 2^x = 2^6`

`=> x = 6`

Vậy, `x = 6`

______

`2^x - 26 = 6`

`=> 2^x = 6 + 26`

`=> 2^x = 32`

`=> 2^x = 2^5`

`=> x = 5`

Vậy, `x = 5`

`3^x*3 = 243`

`=> 3^x * 3 = 3^5`

`=> 3^x = 3^5 \div 3`

`=> 3^x = 3^4`

`=> x = 4`

Vậy, `x = 4.`

2x.(4x+3)+5=x.(8x+4)+1

⇒8x²+6x+5=8x²+4x+1

⇒2x=-4

⇒x=-2

\(2x\left(4x+3\right)+5=x\left(8x+4\right)+1\\ \Leftrightarrow8x^2+6x+5=8x^2+4x+1\\ \Leftrightarrow8x^2-8x^2+6x-4x=1-5\\ \Leftrightarrow2x=-4\\ \Leftrightarrow x=-4:2\\ \Leftrightarrow x=-2\)

    Vậy \(x=-2\)

Đáp án C

Đặt:

t = 2 − x + 2 x + 2 ⇔ t 2 = x + 4 + 2 2 − x 2 x + 2 ⇔ x + 2 2 − x 2 x + 2 = t 2 − 4

Với x ∈ − 1 ; 2 ta được:

t ' = − 1 2 2 − x + 1 2 x + 2 = 0 ⇔ x = 1 ⇒ 3 ≤ t ≤ 3

Khi đó bất phương trình trở thành:

t 2 − 4 > m + 4 t ⇔ m < f t = t 2 − 4 t − 4 *

Để (*) có nghiệm trên đoạn 3 ; 3  khi và chỉ khi  m < max 3 ; 3 f t = − 7

3:

a: 3^x*3=243

=>3^x=81

=>x=4

b; 2^x*16^2=1024

=>2^x=4

=>x=2

c: 64*4^x=16^8

=>4^x=4^16/4^3=4^13

=>x=13

d: 2^x=16

=>2^x=2^4

=>x=4

a) \(2^{2x}.2^4=1024\)

\(2^{2x}=1024:2^4\)

\(2^{2x}=1024:16\)

\(2^{2x}=64\)

\(2^{2x}=2^6\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=3\)

vay \(x=3\)

b) \(2.3^x=10.3^{12}+8.27^4\)

\(2.3^x=2.5.3^{12}+2^3.\left(3^3\right)^4\)

\(2.3^x=2.5.3^{12}+2^3.3^{12}\)

\(2.3^x=2.3^{12}.\left(5+2^2\right)\)

\(2.3^x=2.3^{12}.9\)

\(2.3^x=2.3^{12}.3^2\)

\(2.3^x=2.3^{14}\)

\(\Rightarrow x=14\)

vay \(x=14\)

c) \(5^8.25^x+1=5^{17}\)

\(5^8.\left(5^2\right)^x+1=5^{17}\)

\(5^8.5^{2x}+1=5^{17}\)

\(5^{8+2x}=5^{17}-1\)

e) \(\left(2x-4\right)^5=\left(2x-4\right)^3\)

\(\left(2x-4\right)^5-\left(2x-4\right)^3=0\)

\(\left(2x-4\right)\left[\left(2x-4\right)^2-1\right]=0\)

\(\left(2x-4\right)\left(2x-4-1\right)\left(2x-4+1\right)=0\)

\(\left(2x-4\right)\left(2x-5\right)\left(2x-3\right)=0\)

\(\Rightarrow2x-4=0\)hoac  \(\orbr{\begin{cases}2x-5=0\\2x-3=0\end{cases}}\)

\(\Rightarrow2x=4\)hoac \(\orbr{\begin{cases}2x=5\\2x=3\end{cases}}\)

 \(\Rightarrow x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)

              vay \(x=2\)hoac \(\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{3}{2}\end{cases}}\)