Question

[Gia Huy To]

Answer 1

[HT.Phong (9A5)]

a) 

\(32< 2^x< 128\\ =>2^5< 2^x< 2^7\\ =>5< x< 7\\ =>x=6\)

b) 

\(2\cdot16\ge2^x>4\\ =>2\cdot2^4\ge2^x>2^2\\ =>2^5\ge2^x>2^2\\ =>5\ge x>2\\ =>x\in\left\{3;4;5\right\}\)

c) 

\(9\cdot27\le3^x\le243\\ =>3^2\cdot3^3\le3^x\le3^5\\ =>3^5\le3^x\le3^5\\ =>5\le x\le5\\ =>x=5\)

d) 

\(x^{2019}=x\\ =>x^{2019}-x=0\\ =>x\left(x^{2018}-1\right)=0\)

TH1: x = 0

TH2: `x^2018-1=0`

`=>x^2018=1`

`=>x^2018=1^2018`

`=>x=1` hoặc `x=-1`

Answer 2

[Nguyễn Lê Phước Thịnh]

a: \(32< 2^x< 128\)

=>\(2^5< 2^x< 2^7\)

=>5<x<7

mà x là số tự nhiên

nên x=6

b: \(2\cdot16>=2^x>4\)

=>\(2^5>=2^x>2^2\)

=>2<x<=5

mà x là số tự nhiên

nên \(x\in\left\{3;4;5\right\}\)

c: \(9\cdot27< =3^x< =243\)

=>\(243< =3^x< =243\)

=>\(3^x=243=3^5\)

=>x=5

d: \(x^{2019}=x\)

=>\(x\left(x^{2018}-1\right)=0\)

=>\(\left[{}\begin{matrix}x=0\\x^{2018}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{2018}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

e: \(2^{x+1}+4\cdot2^x=3\cdot2^7\)

=>\(2^x\cdot2+4\cdot2^x=6\cdot2^6\)

=>\(6\cdot2^x=6\cdot2^6\)

=>x=6

f: \(2^{2x}+2^{2x+3}=3^2\cdot8^4\)

=>\(2^{2x}+2^{2x}\cdot8=9\cdot8^4\)

=>\(9\cdot2^{2x}=9\cdot2^{12}\)

=>2x=12

=>x=6

g: \(27^{x+1}=9^{x+5}\)

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

=>3(x+1)=2(x+5)

=>3x+3=2x+10

=>3x-2x=10-3

=>x=7

h: \(3^{x+2}+5\cdot3^{x+1}=648\)

=>\(3^x\cdot9+5\cdot3^x\cdot3=648\)

=>\(3^x\cdot24=648\)

=>\(3^x=\dfrac{648}{24}=27=3^3\)

=>x=3