Log of number to a certain base is 7 whereas log of 8 times the original to the base of 2 times to the base is 5.what is the original number
Answers
Answer:
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Step-by-step 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
Answer:
The original number is 128.
Step-by-step explanation:
Given: the log of a number to a specific base is 7. whereas 5 is the log of 8 times the original number to the base of 2 times the original base.
To discover: What was the original number
Solution:
logₐ (n) = 7
=> log n/log a = 7
=> log n = 7. log a =
=> log₂ₐ (8n) = 5
=> log8n/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
=> 5 log 2 - log 8 = 2 log a
=> 2 log a = log 25 minus log 8
=> log 32 - log 8 = 2 log
=> log 2 a = log ( 32 . 8 )
=> log 4 = 2 log a
=> 2 log 2 = 2 log a
=> log 2 = log a
=> a = 2logₐ (n) = 7
=> log₂ n = 7
=> n = 2⁷ = 128
The original number is 128.
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