how to compare

hi Brent
thank you for reply.
below is problem. i know that it is DS problem, so must be posted in DS section. but i am interested in the approach here

Is x > 10^10 ?
(1) x > 2^34
(2)(2) x = 2^35
oa is D the source manhattan CAT

clock60 wrote:hi Brent
thank you for reply.
below is problem. i know that it is DS problem, so must be posted in DS section. but i am interested in the approach here

Is x > 10^10 ?
(1) x > 2^34
(2)(2) x = 2^35
oa is D the source manhattan CAT

(1) 2^34 = 2^ (3.4 × 10) = (2^3.4) ^10, and since a rough estimation could make us believe that 2^3.4 is greater than 10, hence anything more than 2^34 is got to be greater than 10^10. Sufficient

(2) The explanation above in (1) helps us call this too sufficient.

[spoiler]D[/spoiler]

hi sanju09
your solving is too smart for me,( i am sure that it is valid)
i want to replicate with my own words

2^34=2^(3,4*10)=(2^3,4)^10 we need to compare with 10^10

2^3,4=(10+x) where x is any positive number

so (10+x)^10>10^10
and it happens that 2^34>10^10
i see it is genious

but how you guess that 2^3.4>10
it is really 10.556
anyway many many thanks

clock60 wrote:hi sanju09
your solving is too smart for me,( i am sure that it is valid)
i want to replicate with my own words

2^34=2^(3,4*10)=(2^3,4)^10 we need to compare with 10^10

2^3,4=(10+x) where x is any positive number

so (10+x)^10>10^10
and it happens that 2^34>10^10
i see it is genious

but how you guess that 2^3.4>10
it is really 10.556
anyway many many thanks

2^3.4 = 4^1.7, which is faintly less than the square of 4, not enough that could take 16 below 10 in any case, it’s a sheer ballpark figure to make persuaded about something else rather than the exact value.