Answer :
[tex]n^{200}<5^{300}\\\\(n^2)^{100}<(5^3)^{100}\ \ \ \Rightarrow\ \ \ n^2<5^3\ \ \ \Rightarrow\ \ \ n^2<125\\\\n\in I\ \ \ \Rightarrow\ \ \ n< \sqrt{125} \ \ \ and\ \ \ \sqrt{125} =5 \sqrt{5} \approx11.18\\\\Ans.\ the\ largest\ integer\ n\ is\ 11[/tex]
Let us take logarithms on both sides.
200 log n < 300 log 5
So log n < 3/2 log 5
log n < log 5 power 3/2
n < 5 power 3/2
n < square root (5³) = √125
so n = 11
200 log n < 300 log 5
So log n < 3/2 log 5
log n < log 5 power 3/2
n < 5 power 3/2
n < square root (5³) = √125
so n = 11
