Many Logs Inequality

Algebra Level 3

$\log 1 + \log 2 + \log 3 + \log 4 + \log 5 < 2 \log N$

What is the smallest integer $N$ such that the above inequality is true?

120 15 10 11

2 solutions

Mateus Gomes
Mar 5, 2016

$\log 1 + \log 2 + \log 3 + \log 4 + \log 5 < 2 \log N$ $\log 120< 2 \log N$ $120<N^2$ $smallest~~integer~~N=11$

I missed the 2 :c

Mehul Arora - 5 years, 3 months ago

Nicola M.
Mar 5, 2016

We can solve like this: log1+log2+log3+log4+log5<2logN For the properties of logs, we know that: log(a)+log(b)=log(a b) and: n log(a)=log(a^n)

So: log(5x4x3x2x1)=log(5!)<log(N^2)

We can now say that: 5!<N^2 120<N^2 N=√120=10.95=11