\(\frac{2^{4-x}}{16^5}=32^6\)

\(\Rightarrow\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)

\(\Rightarrow\frac{2^{4-x}}{2^{20}}=2^{30}\)

\(\Rightarrow2^{4-x}=2^{30}.2^{20}\)

\(\Rightarrow2^{4-x}=2^{50}\)

\(\Rightarrow4-x=50\)

\(\Rightarrow x=-46\)

64 . 4^x = 168

<=> 43. 4^x = 416

=> 3 + x = 16

<=> x = 13

Vậy x = 13

2^x.16² = 1024

<=> 2^x. 2⁸ = 210

=> x + 8 = 10

<=> x = 2

Vậy x = 2

b: Ta có: \(2^x\cdot16^2=1024\)

\(\Leftrightarrow2^x\cdot2^8=2^{10}\)

\(\Leftrightarrow x+8=10\)

hay x=2

Dễ thế mà ko biết làm

a) x=10

b) x=1

a)x=-2

b)x=-102

c)x=123

d)x=10

e)x=-48

256 x 6=512 x 3

512 x 3=256 x 6

999 x 2>1000

512 x 5>1024

1024<512 x 5

1024=512 x 2

512=256 x 2

256=128 x 2

128=64 x 2

64=32 x 2

32=16 x 2

Đúng rồi

\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)

\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)

\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)

\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)

\(\Rightarrow3x=1023\)

\(\Rightarrow x=341\)

Lời giải:

$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$

$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$

$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$

$\frac{3x}{1024}=\frac{1023}{1024}$

$\Rightarrow 3x=1023$

$\Rightarrow x=341$