prove that
1log2N+1log3N+1log4N+⋯+1log100N=1log100!N
how many real roots
sinx=logx
1.
From the definition of logarithm, we obtain:
1logba=logab
The equation can therefore be written in the form:
logn2+logn3+⋯logn100+=logn(2⋅3⋯100)
from which the desired conclusion immediately follows.
2.
https://www.desmos.com/calculator/vp8etonnna
First, if sinx=logx, then x≤10 (inasmuch as sinx≤1.
Since 2⋅2π>10, the interval on the x-axis between x = 0 and x = 10 contains one complete period of the sine curve plug part of a second period. The graph of logx intersects the first wave of the sine curve at precisely one point. Futher, since 2Π+Π2<10, than at the point x=5Π2, we have sinx=1>logx, which means that the graph of log x intersects the first half of the second positive wave of sin x. Since, at x=10,logx=1>sinx, the graph intersect this second wave another time. Therefore we conclude that the equation sinx=logx, has exactly three roots.
Wow! I haven't done these in SO long! Thanks for bringing me back to the more beautiful realm of mathematics.
I hope this helped,
Gavin
Suppermanaccz,
It is good that you have stated that you do not understand.
BUT
You have not even said which answer you do not understand. There are 2 different question!
You need to say WHAT question and also try to say where you get lost.
If you do not understand the first line then say so.
If you do understand the first line then tell GYanggg what you do not understand!