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\(8^2.32^4=2^{2.3}.2^{5.4}=2^6.2^{20}=2^{26}\)
\(64^3.256^2=2^{6.3}.2^{8.2}=2^{18}.2^{16}=2^{34}\)
\(4^{21}=\left(2^2\right)^{21}\)\(=2^{42}\)
\(64^7=\left(2^6\right)^7=2^{42}\)
Vì \(2^{42}=2^{42}\)
\(\Rightarrow4^{21}=64^7\)
\(4^{21}=\left(2^2\right)^{21}=2^{42}\)
\(64^7=\left(2^6\right)^7=2^{42}\)
Vậy \(4^{21}=64^7\)
Bài 1:
a) \(8^5\cdot8^2=8^7\)
b) \(9^3\cdot3^2=\left(3^2\right)^3\cdot3^2=3^6\cdot3^2=3^8\)
c) \(2^7\cdot5^7=10^7\)
d) \(27^6:3^3=\left(3^3\right)^6:3^3=3^{18}:3^3=3^{15}\)
Bài 2:
a) \(x^6:x^3=125\)
\(\Rightarrow x^3=125\)
\(\Rightarrow x=5\)
b) \(x^{20}=x\)
\(\Rightarrow x^{20}-x=0\)
\(\Rightarrow x\left(x^{19}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{19}-1=0\Rightarrow x=1\end{matrix}\right.\)
c) \(3^x\cdot3=243\)
\(\Rightarrow3^x=81\)
\(\Rightarrow x=4\)
d) \(2x-138=2^3\cdot3^2\)
\(\Rightarrow2x-138=72\)
\(\Rightarrow2x=200\)
\(\Rightarrow x=100\)
Giải:
Bài 1:
a) \(8^5.8^2=8^{5+2}=8^7\)
b) \(9^3.3^2=3^6.3^2=3^{6+2}=3^8\)
c) \(2^7.5^7=\left(2.5\right)^7=10^7\)
d) \(27^6:3^3=3^{18}:3^3=3^{18-3}=3^{15}\)
Bài 2:
a) \(x^6:x^3=x^{6-3}=x^3=125\)
\(\Leftrightarrow x=5\)
b) \(x^{20}=x\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=1\end{matrix}\right.\)
c) \(3^x.3=243\)
\(\Leftrightarrow3^{x+1}=243\)
\(\Leftrightarrow3^{x+1}=3^5\)
\(\Leftrightarrow x+1=5\Leftrightarrow x=4\)
d) \(2.x-138=2^3.3^2\)
\(\Leftrightarrow2.x-138=8.9\)
\(\Leftrightarrow2.x-138=72\)
\(\Leftrightarrow2.x=72+138\)
\(\Leftrightarrow2.x=210\Leftrightarrow x=105\)
Chúc bạn học tốt!
Ta có:
12^8=(3.2^2)^8=3^8.2^16
27^16.16^9=(3^3)^16.(2^4)^9=3^48.2^36
<=>12^8<27^16.16^9
b) \(\Leftrightarrow18x-288=27.4+8.9-144\)
\(\Leftrightarrow18x=108+72-144+288=324\)
\(\Leftrightarrow x=18\)
\(32^{4}=(2^{5})^{4}=2^{20}\)
\(64^{12}=(2^{6})^{12}=2^{72}\)
Vì `20 < 72=>` \(2^{20} < 2^{72}\)
Hay \(32^{4} < 64^{12}\)