数字有一个可爱的结果 理论表明对于 的乘积足够大 小于 n 的素数小于或 等于 e^n 并且随着 n 的增长， 这成为一个紧界（即 素数的乘积之比 随着 n 的增长，e^n 接近 1）。
计算大数的乘积 质数的数量可能会导致非常 数量很大，这可能会 导致我们的计算出现问题。 [注：这就是约翰所指的 到]所以我们可以将a的乘积转换为 一组素数化为总和 应用素数的对数 这两个部分的对数 推测。在这种情况下， 上述猜想简化为断言 所有的对数之和 小于 n 的素数小于 n， 随着 n 的增长，这个比率 n 的总和接近 1。
然而
，鉴于这些陈述，我不确定如何应用它们，即
我们如何从这里开始：
2 x 3 x 5 <= e^7
到
“正在申请 这两个部分的对数 推测。”
编辑 2
明白了...
2 x 3 x 5 <= e^7
知道对数与幂相反，我们可以说：
log(2x3x5) <= 7
这是也与：
log(2)+log(3)+log(5) <= 7
只有当 n（在本例中为 7）变大（即第 1000 个质数或更高）时才开始显示其“值”
thanks for that OrangeDog and John!
re benefit of introducing logs, OrangeDog is right indeed. It is specific to an exercise from an MIT OpenCourse class. Here's the full details:
There is a cute result from number theory that states that for sufficiently large n the product of the primes less than n is less than or equal to e^n and that as n grows, this becomes a tight bound (that is, the ratio of the product of the primes to e^n gets close to 1 as n grows).
Computing a product of a large number of prime numbers can result in a very large number, which can potentially cause problems with our computation. [note: this is what John was referring to] So we can convert the product of a set of primes into a sum of the logarithms of the primes by applying logarithms to both parts of this conjecture. In this case, the conjecture above reduces to the claim that the sum of the logarithms of all the primes less than n is less than n, and that as n grows, the ratio of this sum to n gets close to 1.
EDIT
given these statements i am, however, unsure about how to apply them i.e.
how do we go from here:
2 x 3 x 5 <= e^7
to
"applying logarithms to both parts of this conjecture."
EDIT 2
got it...
2 x 3 x 5 <= e^7
knowing that logarithms are the opposite of powers we can say:
log(2x3x5) <= 7
which is also the same as:
log(2)+log(3)+log(5) <= 7
this only starts to show its "value" when n (in this case 7) gets larger i.e. the 1000th prime or higher
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log(a * b) = log(a) + log(b)log(a * b) = log(a) + log(b)感谢 OrangeDog 和约翰！
关于引入日志的好处，OrangeDog 确实是对的。它特定于麻省理工学院开放课程课程的练习。以下是完整的详细信息：
然而
，鉴于这些陈述，我不确定如何应用它们，即
我们如何从这里开始：
2 x 3 x 5 <= e^7
到
编辑 2
明白了...
2 x 3 x 5 <= e^7
知道对数与幂相反，我们可以说：
log(2x3x5) <= 7
这是也与：
log(2)+log(3)+log(5) <= 7
只有当 n（在本例中为 7）变大（即第 1000 个质数或更高）时才开始显示其“值”
thanks for that OrangeDog and John!
re benefit of introducing logs, OrangeDog is right indeed. It is specific to an exercise from an MIT OpenCourse class. Here's the full details:
EDIT
given these statements i am, however, unsure about how to apply them i.e.
how do we go from here:
2 x 3 x 5 <= e^7
to
EDIT 2
got it...
2 x 3 x 5 <= e^7
knowing that logarithms are the opposite of powers we can say:
log(2x3x5) <= 7
which is also the same as:
log(2)+log(3)+log(5) <= 7
this only starts to show its "value" when n (in this case 7) gets larger i.e. the 1000th prime or higher