Answer :
Let x = the number 2 must be raised to to make 100
Let y = the number 2 must be raised to to make 200
[tex]2^x = 100[/tex]
[tex]\log 2^x = \log 100[/tex]
[tex]x \log 2 = \log{100}[/tex]
[tex]x = \frac{\log 100}{\log 2} [/tex]
[tex]x \approx 6.64[/tex]
[tex]2^y = 200[/tex]
[tex]\log 2^y = \log 200[/tex]
[tex]y \log 2 = \log{200}[/tex]
[tex]y = \frac{\log 200}{\log 2} [/tex]
[tex]y \approx 7.64[/tex]
So the integer powers of 2 that would fall between 100 and 200 are between 6.64 and 7.64 - 7 is the only one.
Following a similar method for the second question:
Let z = the number 2 must be raised to to make 1000
[tex]2^z = 1000[/tex]
[tex]\log 2^z = \log 1000[/tex]
[tex]z \log 2 = \log 1000[/tex]
[tex]z = \frac{\log 1000}{\log 2} [/tex]
[tex]z \approx 9.97[/tex]
So [tex]2^{10}[/tex] is the closest power of 2 to 1000.
Let y = the number 2 must be raised to to make 200
[tex]2^x = 100[/tex]
[tex]\log 2^x = \log 100[/tex]
[tex]x \log 2 = \log{100}[/tex]
[tex]x = \frac{\log 100}{\log 2} [/tex]
[tex]x \approx 6.64[/tex]
[tex]2^y = 200[/tex]
[tex]\log 2^y = \log 200[/tex]
[tex]y \log 2 = \log{200}[/tex]
[tex]y = \frac{\log 200}{\log 2} [/tex]
[tex]y \approx 7.64[/tex]
So the integer powers of 2 that would fall between 100 and 200 are between 6.64 and 7.64 - 7 is the only one.
Following a similar method for the second question:
Let z = the number 2 must be raised to to make 1000
[tex]2^z = 1000[/tex]
[tex]\log 2^z = \log 1000[/tex]
[tex]z \log 2 = \log 1000[/tex]
[tex]z = \frac{\log 1000}{\log 2} [/tex]
[tex]z \approx 9.97[/tex]
So [tex]2^{10}[/tex] is the closest power of 2 to 1000.
