18hun699
18hun699 18hun699
  • 09-10-2017
  • Mathematics
contestada

use the fact that 10^3 = 2^10 to mentally find a vaule of n for which 2^n > 10^50 & 10^n > 2^300

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LammettHash
LammettHash LammettHash
  • 09-10-2017
[tex]2^n>10^{50}=10^{3\cdot16+2}=10^2(10^3)^{16}=10^2(2^{10})^{16}=5^22^{162}[/tex]
[tex]2^{n-162}>25[/tex]

The least power of 2 that exceeds 25 is [tex]2^5=32[/tex], so we have

[tex]2^{n-162}=2^N>25\implies N=5\implies n-162=5\implies n=167[/tex]

- - -

[tex]10^n>2^{300}=(2^{10})^{30}=(10^3)^{30}=10^{90}[/tex]

The least integer [tex]n[/tex] that satisfies this inequality would clearly be [tex]n=91[/tex].
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