In `log_2 3.log_34.log_45...log_n(n+1)=10` find n

Solution:

We can write it as:

`log_23.\frac{log_24}{log_23}.\frac{log_25}{log_24}...\frac{log_2(n+1)}{log_2n}=10`

`\Rightarrow log_2(n+1)=10`

`\Rightarrow n+1=2^10`

`\Rightarrow n=1023`

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