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a) 3^n = 243 => 3^n = 3^5 => n = 5
b) 2^n = 256
=> 2^n = 2^8
=> n = 8
c) 64.4^n = 4^5
=> 4^3.4^n = 4^5
=>4^3. 4^n= 4^2 .4^3
=> n = 2
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a) \(3^n=243\)
\(\Leftrightarrow3^n=3^5\)
\(\Leftrightarrow n=5\left(TM\right)\)
Vậy \(n=5\)
b) \(2^n=256\)
\(\Leftrightarrow2^n=n^8\)
\(\Leftrightarrow n=8\left(TM\right)\)
Vậy \(n=8\)
c) \(3^{1234}=\left(3^2\right)^{617}=9^{617}\)
\(2^{1851}=\left(2^3\right)^{617}=8^{617}\)
Vì \(9^{617}>8^{617}\Leftrightarrow3^{1234}>2^{1851}\)
d) \(6^{30}=\left(6^2\right)^{15}=36^{15}\)
Vì \(36^{15}>12^{15}\Leftrightarrow6^{30}>12^{15}\)
1.
a, \(3^n=243\)
\(3^n=3^5\)
\(\Rightarrow n=5\)
b, \(2^n=256\)
\(2^n=2^8\)
\(\Rightarrow n=8\)
2.
a,\(3^{1234}\)và \(2^{1851}\)
\(3^{1234}=\left(3^2\right)^{617}=9^{617}\)
\(2^{1851}=\left(2^3\right)^{617}=8^{617}\)
Ta thấy \(9^{617}>8^{617}\Rightarrow3^{1234}>2^{1851}\)
b, \(6^{30}\)và \(12^{15}\)
\(6^{30}=\left(6^2\right)^{15}=36^{15}\)
Ta thấy \(36^{15}=12^{15}\Rightarrow6^{30}>12^{15}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
B1:
a) 3n = 243
3n = 35
\(\Rightarrow\)n = 5
b) 2n = 256
2n = 28
=> n = 8
Câu 1:
a)\(3^n=243\)
Ta có:\(3^n=3^5\Rightarrow n=5\)
b)\(2^n=256\)
Ta có:\(2^n=2^8\Rightarrow n=8\)
Câu 2:
a)31234 và 21851
Ta có:\(3^{1234}=\left(3^2\right)^{617}=9^{617}\)
\(2^{1851}=\left(2^3\right)^{617}=8^{617}\)
Vì \(8^{617}< 9^{617}\)
Vậy \(2^{1851}< 3^{1234}\)
b)630 và 1215
Ta có:\(6^{30}=\left(6^2\right)^{15}=36^{15}\)
Vì \(12^{15}< 36^{15}\)
Vậy \(12^{15}< 6^{30}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(256< 2^n< 1024\)
\(\Leftrightarrow2^8< 2^n< 2^{10}\)
\(\Rightarrow8< n< 10\)
\(\Rightarrow n=10\)
Vậy \(n=10\)
b) Ta có: \(27< 3^n< 243\)
\(\Leftrightarrow3^3< 3^n< 3^5\)
\(\Rightarrow3< n< 5\)
\(\Rightarrow n=4\)
Vậy \(n=4\)
c) Ta có: \(16< 4^n< 256\)
\(\Leftrightarrow4^2< 4^n< 4^4\)
\(\Rightarrow2< n< 4\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
d) Ta có: \(125< 5^n< 3125\)
\(\Leftrightarrow5^3< 5^n< 5^5\)
\(\Rightarrow3< n< 5\)
\(\Rightarrow n=4\)
Vậy n=4
a) \(256< 2^n< 1024\)
Ta có : \(2^8< 2^n< 2^{10}\)
Vậy n = 9
b) \(27< 3^n< 243\)
Ta có : \(3^3< 3^n< 3^5\)
Vậy n = 4
c) \(16< 4^n< 256\)
Ta có : \(4^2< 4^n< 4^4\)
Vậy n = 3
d) \(125< 5^n< 3125\)
Ta có : \(5^3< 5^n< 5^5\)
Vậy n = 4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 27. 3n=243. 3n
=243:27
=9. có 3n
=9=32.
=>3n=32. => n=2.
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