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64 . 4x = 168
<=> 43. 4x = 416
=> 3 + x = 16
<=> x = 13
Vậy x = 13
2x.162 = 1024
<=> 2x. 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
1. \(2^x-26=6\)
\(\Rightarrow2^x=6+26\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
2. \(64\cdot4^x=16^8\)
\(\Rightarrow4^3\cdot4^x=4^{16}\)
\(\Rightarrow4^x=4^{16}:4^3\)
\(\Rightarrow4^x=4^{13}\)
\(\Rightarrow x=13\)
3. \(\left(2x-1\right)^4=16\)
\(\Rightarrow\left(2x-1\right)^4=2^4\)
\(\Rightarrow2x-1=2\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\dfrac{3}{2}\)
4. \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Ta có: \(64\cdot4^x=168\)
\(\Leftrightarrow4^x=\dfrac{21}{8}\)
hay \(x\in\varnothing\)
Bài làm :
\(4^n=64.4\)
\(\Rightarrow4^n=256\)
\(\Rightarrow4^n=4^4\)
\(\Rightarrow n=4\)
Vậy n = 4 .
Học tốt
\(4^n=64.4\)
\(4^n=256\)
\(4^n=4^4\)
\(n=4\)
vậy \(n=4\)
\(2^x+5\cdot2^{x+2}=168\)
=>\(2^x+20\cdot2^x=168\)
=>\(21\cdot2^x=168\)
=>\(2^x=8=2^3\)
=>x=3
\(2^x+5.2^{x+2}=168\\ \Leftrightarrow2^x.\left(1+5.2^2\right)=168\\ \Leftrightarrow2^x.21=168\\ \Leftrightarrow2^x=\dfrac{168}{21}=8=2^3\\ Vây:x=3\)
2x+5.2x+2=168⇔2x.(1+5.22)=168⇔2x.21=168⇔2x=16821=8=23
Vây:x=3
Ta có: \(64\cdot4^x=168\)
\(\Leftrightarrow4^x=\dfrac{21}{8}\)
hay \(x\in\varnothing\)