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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
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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
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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

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