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Mathematics 21 Online
OpenStudy (anonymous):

True or False... There is some value x > 2 for which x^100 <1.01^x

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

eventually exponential always beats out polynomial

OpenStudy (anonymous):

I kinda don't get what the question is asking..

OpenStudy (anonymous):

is there a number that makes \((1.01)^x\) greater that \(x^{100}\)

OpenStudy (anonymous):

certainly not true if \(x=50\) because \((1.01)^{50}\) is less than \(50^{100}\)

OpenStudy (anonymous):

the numbers that make it bigger will be very very large,

OpenStudy (anonymous):

so 100 log x and x log 1.01

OpenStudy (anonymous):

yeah why not? solve \[x\log(1.01)>100\log(x)\]

OpenStudy (anonymous):

\[\log(1.01)\] is a very very small number (close to zero) but is it positive and since \(x\) grows much faster than \(\log(x)\) at some point \[x\log(1.01)\] will be greater than \(100\log(x)\)

OpenStudy (anonymous):

so like a very large number? so maybe plug in 10^10 for x?

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