+0

0
112
2

Suppose that $\frac{\log_3 2}{\log_2 9} \cdot \frac{\log_3 4}{\log_2 27} \cdot \frac{\log_3 8}{\log_2 81} \cdots \frac{\log_3 2^{100}}{\log_2 3^{101}} = \frac{(\log_3 2)^m}{n}$, where $m$ and $n$ are integers. Find $m+n$.

Jun 1, 2022

#1
+505
+1

Might as well make this easier for others to read.

Suppose that $$\frac{\log_3 2}{\log_2 9} \cdot \frac{\log_3 4}{\log_2 27} \cdot \frac{\log_3 8}{\log_2 81} \cdots \frac{\log_3 2^{100}}{\log_2 3^{101}} = \frac{(\log_3 2)^m}{n}$$, where m and n are integers. Find m+n.

Jun 1, 2022
#2
0

9.81477439992×10^-43==0.6309297536^ m/n, solve for m

m==200,   n==101

m  +  n ==200  +  101 ==301

Note: there maybe other integer solution for m and n.

Jun 1, 2022