a) 2^x . 4 = 128

<=> 2^x = 32 

<=> 2^x = 2⁵

<=> x = 5

b) x¹⁵ = x¹

<=> x¹⁵ - x = 0

<=> x(x¹⁴ - 1) = 0

<=> \(\orbr{\begin{cases}x=0\\x^{14}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{14}=1^{14}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

c) (2x + 1)³ = 125

<=> (2x + 1)³ = 5³

<=> 2x + 1 = 5

<=> 2x = 4

<=> x = 2

d) (x - 5)⁴ = (x - 5)⁶

<=> (x - 5)⁶ - (x - 5)⁴ = 0

<=> (x - 5)⁴[(x - 5)² - 1] = 0

<=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}}\)

Khi (x - 5)⁴ = 0 => x - 5 = 0 => x = 5

Khi (x - 5)² - 1 = 0 <=> (x - 5)² = 1² <=> \(\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

a, 2x . 4 = 128

=> 2x = 128 : 4 = 32

=> x = 32 : 2 = 16

Vậy x = 16

b, x15 = x 1 => Sai đề

c, (2x + 1)³ = 125

=> ( 2x + 1 ) = 5³

=> 2x + 1 = 5 

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

Ta có : (x - 5)⁴ = (x - 5)⁶

=>  (x - 5)⁴ - (x - 5)⁶ = 0

=> (x - 5)⁴ (1 - (x - 5)²) = 0

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)=0\\\left(x-5\right)^2=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x-5=1;-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)

Vậy x = {4;5;6}

Tìm x thuộc N:

2x.4=128

2x    =128:4

2x     =32

 x       =32:2

 x       =16

b.x.15=x

   x thỏa mãn điều kiện:0,1

c.(2x+1)³=125

   (2x+1)³=5³

==>2x+1=5

      2x    =5-1

      2x     =4

        x      =4:2=2

phần d mk chưa hiểu lắm

+) Cấm ghi "nhanh lên nhé ai giải được thì tk cho"

a) 2x.4 = 128

    2x    = 128 : 4

    2x    = 32

    2x    = 2⁵

b) x15 = x

    => x = 0

c) (2x + 1)³ = 125

   (2x + 1)³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 2

d) (x - 5).4 = (x - 5).6

=> x = 0

a,   2^x . 4 = 128

           2^x = 128 : 4

           2^x = 32

           2^x = 2⁵

   =>     x = 5

b,   x¹⁵ = x

=> x = 1 hoặc x = -1 hoặc x = 0

c,   ( 2x + 1 )³ = 125

=>  ( 2x + 1 )³ = 5³

=>    2x + 1     = 5

=>          2x     = 5 - 1

=>          2x     = 4

=>            x     = 4 : 2

=>            x     = 2

a;2^x*4 =128                        b;x^15=x                            c;(2x+1)^3=125                x^10=1^x

2^x    =128/4                        ta có 3 trường hợp             (2x+10)^3=5^3           suy ra1^10=1^10

2^x    =32                     nếu x=0 thì x^15=x=0                 2x+10=5

mà 32 =2^5                         x=1 thì x^15=x=1                 2x     =5-10

suy ra 2^x =2^5                   x=-1 thì x^15=x=-1               2x     =-5

vậy x=5                                                                         x      =-5/2          

                                                                                    x       =-2.5

\(a,2^x.4=128\\2^x.2^2=2^7\\ 2^x=\dfrac{2^7}{2^2}=2^{7-2}=2^5\\ Vậy:x=5\\ ----\\ b,\left(2x+1\right)^3=125=5^3\\ \Rightarrow 2x+1=5\\ 2x=5-1=4\\ x=\dfrac{4}{2}=2\\ ----\\ c,2x-2^6=6\\ 2x=6+2^6=6+64\\ 2x=70\\ x=\dfrac{70}{2}=35\\ ----\\ d,49.7^x=2401\\ 7^x=\dfrac{2401}{49}=49=7^2\\ Vậy:x=2\)

a) x=5

b) x=0 hoặc 1

c) x=2

d) x=6 hoặc 5

e) x=0 hoặc 1

f) x=8

a, 2^x . 4 = 128

2^x = 128 : 4

2^x = 32

2^x = 2⁵

=> x = 5

b, x¹⁵ = x¹

=> x¹⁵ - x = 0

x . ( x¹⁴ - 1 ) = 0

=> x = 0 hoặc x¹⁴ - 1 = 0 

=> x = 0 hoặc x = 1

c, (2x + 1)³ = 125

( 2x + 1 )³ = 5³

=> 2x + 1 = 5

=> 2x = 5 - 1

=> 2x = 4

=> x = 4 : 2

=> x = 2

d, (x – 5)⁴ = (x - 5)⁶

=> ( x - 5 )⁶ - ( x - 5 )⁴ = 0

=> ( x - 5 )⁴ . [ ( x - 5 )² - 1 ] = 0

=> \(\orbr{\begin{cases}\left(x-5\right)^4=0\\\left(x-5\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}}\)

e, x¹⁰ = x

x¹⁰ - x = 0

x . ( x⁹ - 1 ) = 0

\(\Rightarrow\orbr{\begin{cases}x=0\\x^9-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

f, (2x -15)⁵ = (2x -15)³

( 2x - 15 )⁵ - ( 2x - 15 )³ = 0

( 2x - 15 )³ . [ ( 2x - 15 )² - 1 ] = 0

\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=1\end{cases}\Rightarrow}\orbr{\begin{cases}x\text{ không tồn tại}\\x=8\end{cases}}}\)

a) \(\left(x-1\right)^3=8=2^3\)

\(x-1=2\)

\(x=2+1=3\)

b) \(7^{2x-6}=49=7^2\)

\(2x-6=2\)

\(2x=6+2=8\)

\(x=8:2=4\)

c) \(\left(2x-14\right)^7=128=2^7\)

\(2x-14=2\)

\(2x=14+2=16\)

\(x=16:2=8\)

d) \(x^4\cdot x^5=5^3\cdot5^6=5^4\cdot5^5\)

\(x=5\)

e) \(3\cdot\left(x+2\right):7\cdot4=120\)

\(x+2=120:3\cdot7:4\)

\(x+2=70\)

\(x=70-2=68\)

Lời giải:

a. $(x-1)^3=8=2^3$
$\Rightarrow x-1=2$

$\Rightarrow x=3$

b. $7^{2x-6}=49=7^2$
$\Rightarrow 2x-6=2$

$\Rightarrow 2x=8$

$\Rightarrow x=4$

c. $(2x-14)^7=128=2^7$

$\Rightarrow 2x-14=2$

$\Rightarrow 2x=16$

$\Rightarrow x=18$

d.

$x^4.x^5=5^3.5^6$

$x^9=5^9$

$\Rightarrow x=5$

e. 

$3(x+2):7=120:4=30$

$3(x+2)=30.7=210$

$x+2=210:3=70$

$x=70-2=68$