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((log^2)_25 to 125) - ((log^2)_25 to 5)?
1 Answer
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- L. E. GantLv 74 years ago
Not quite sure of your notations...
but I figure it's
(log to the base 25 of 125) squared - (log to the base 25 of 5) squared
so:
(log_25(125))^2 - (log_25(5))^2 .... difference of squares factorisation
= (log_25(125) + log_25(5)) (log_25(125) - log_25(5))
= log_25(625) * log_25(25)
= 2log_25(25) * 1
= 2*1
= 2
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