3:
a: 4^2x+1=64
=>2x+1=3
=>2x=2
=>x=1
b; =>2x-3=3
=>2x=6
=>x=3
c: =>2(x-1)^3=16
=>(x-1)^3=8
=>x-1=2
=>x=3
d: =>(2x+1)^2*3=3
=>(2x+1)^2=1
=>2x+1=1 hoặc 2x+1=-1
=>x=-1 hoặc x=0
Bài 1:
a) \(\left(3^{14}\cdot15-3^{14}\cdot6\right):3^7\)
\(=3^{14}\cdot\left(15-6\right):3^7\)
\(=3^{14-7}\cdot9\)
\(=3^7\cdot3^2\)
\(=3^{7+2}\)
\(=3^9\)
b) \(\left(2^{15}\cdot9+2^{15}\cdot7\right):2^9\)
\(=2^{15}\cdot\left(9+7\right):2^9\)
\(=2^{15-9}\cdot16\)
\(=2^6\cdot2^4\)
\(=2^{6+4}\)
\(=2^{10}\)
Bài 2:
a) Ta có: \(16^5=\left(2^4\right)^5=2^{20}\)
Mà: \(2^{20}>2^{19}\)
\(\Rightarrow16^5>2^{19}\)
b) Ta có:
\(5^{24}=\left(5^3\right)^8=125^8\)
\(4^{32}=\left(4^4\right)^8=256^8\)
Mà: \(125^8< 256^8\)
\(\Rightarrow5^{^{24}}< 4^{32}\)
