a) 2^x.4 = 128

    2^x    = 128 : 4

    2^x    = 32

    2^x    = 2⁵

=>  x = 5

b) x¹⁵ = x

=> x có hai trường hợp : x = 1 hoặc x = 0

Vì 1¹⁵ = 1 ; 0¹⁵ = 0

Nên x = 1 hoặc 0

c) 16^x < 128

=> \(x\in\left\{0;1\right\}\)

\(a.2^x.4=128\)

\(\Rightarrow2^x=32\)

\(\Rightarrow x=5\)

\(b.x^{15}=x\)

\(\Rightarrow x\in \left\{1;0\right\}\)

Tương tự

a,  2 x . 4 = 128

=>  2 x = 128 : 4

=>  2 x = 32

=>  2 x = 2 5

=> x = 5

Vậy x = 5

b,  x 15 = x

=> x = 1 hoặc x = 0

Vì  1 15 = 1 ;   0 15 = 0

c,  x - 5 4 = x - 5 6

=> x – 5 = 1 hoặc x – 5 = 0

=> x = 6 hoặc x = 5

Vậy x = 6 hoặc x = 5

d,  x 2018 = x 2

=> x = 1 hoặc x = 0

Vì  1 2018 = 1 2 = 1 ;   0 2018 = 0 2 = 0

A=7 B=1 OR =0 C=2 D=4.5.6

a: \(2^{x}\cdot4=128\)

=>\(2^{x}=\frac{128}{4}=32=2^5\)

=>x=5

b: \(x^{15}=x\)

=>\(x^{15}-x=0\)

=>\(x\left(x^{14}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{14}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{14}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)

c: \(\left(2x+1\right)^3=125\)

=>\(\left(2x+1\right)^3=5^3\)

=>2x+1=5

=>2x=5-1=4

=>\(x=\frac42=2\)

d: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-5-1\right)\left(x-5+1\right)=0\)

=>\(\left(x-5\right)^4\cdot\left(x-6\right)\left(x-4\right)=0\)

=>\(\left[\begin{array}{l}x-5=0\\ x-6=0\\ x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=6\\ x=4\end{array}\right.\)

b)

\(5^x.5^{x+1}.5^{x+2}=5^x.5^x.5.5^x.5^2=5^{x+x+x+1+3}=5^{3x+3}\le10^{18}:2^{118}\)

\(=>5^{3x+3}\le5^{18}=>3x+3\le18=>x\le5=>x\in\left\{0;1;2;3;4;5\right\}\)

**** bn, câu a tự lm nhé

\(128^4=\left(2^7\right)^4=2^{28}=\left(2^5\right)^{5,6}=32^{5,6}\)

theo đề ta có

\(32^x< 128^4=32^{5,6}\Rightarrow x\in\left\{1;2;3;4;5\right\}\)