Question
$$\text { Show that } \frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\cdots \cdots+\frac{1}{\log _{43} n}=\frac{1}{\log _{43 !} n} \text { . }$$
Step 1
Applying this property to the given expression, we get: $$ \frac{1}{\log _{2} n}+\frac{1}{\log _{3} n}+\cdots \cdots+\frac{1}{\log _{43} n} = \log_n 2 + \log_n 3 + \cdots + \log_n 43 $$ Show more…
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