log of a number to a certain base is 7 whereas log of 8 times the original number
Answers
Given : log of number to a certain base is 7. whereas log of 8 times the original number to the base of 2 times the original base is 5.
To Find : What is the original number?
Solution:
logₐ (n) = 7
=> log n / log a = 7
=> log n = 7 . log a
log₂ₐ (8n) = 5
=> log 8n / log2a = 5
=> log 8n = 5 . log 2a
=> log 8 + log n = 5 . log 2a
=> log 8 + 7 . log a = 5 . log 2a
=> log 8 + 7 . log a = 5 log 2 + 5 log a
=> 2 log a = 5 log 2 - log 8
=> 2 log a = log 2⁵ - log 8
=> 2 log a = log 32 - log 8
=> 2 log a = log ( 32 . 8 )
=> 2 log a = log 4
=> 2 log a = 2 log 2
=> log a = log 2
=> a = 2
logₐ (n) = 7
=> log₂ n = 7
=> n = 2⁷ = 128
Original number is 128
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Answer:
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Explanation:
logₐ (n) = 7
=> log n / log a = 7
=> log n = 7 . log a
log₂ₐ (8n) = 5
=> log 8n / log2a = 5
=> log 8n = 5 . log 2a
=> log 8 + log n = 5 . log 2a
=> log 8 + 7 . log a = 5 . log 2a
=> log 8 + 7 . log a = 5 log 2 + 5 log a
=> 2 log a = 5 log 2 - log 8
=> 2 log a = log 2⁵ - log 8
=> 2 log a = log 32 - log 8
=> 2 log a = log ( 32 . 8 )
=> 2 log a = log 4
=> 2 log a = 2 log 2
=> log a = log 2
=> a = 2
logₐ (n) = 7
=> log₂ n = 7
=> n = 2⁷ = 128
Original number is 128