### Logarithms.

is the inverse function to exponentation, as it answers the question, how many of one number do we multiply to get another number.
log_b a = (log a)/(log b)

Subjects > Mathematics > Logarithms

Question 03:

Solve: log_2 8 - log_(1/2) 8

###### Solution:
Given,
log_2 8 - log_(1/2) 8
we cannot use quotient rule here, since bases are not equal in each term
Hence,
we know that,
log_b a = (log a)/(log b)

⇒ log_2 (2^3) - (log 8)/(log (1/2))

But we know that, 1/n = n^-1

⇒ 3log_2 2 - (log 8)/(log 2^-1)

⇒ 3 - (log 8)/(-log 2)

⇒ 3 - (-log 8)/(log 2)

⇒ 3 + log_2 8
⇒ 3 + log_2 2^3
⇒ 3 + 3log_2 2
⇒ 3 + 3 * 1 = 6
∴ log_2 8 - log_(1/2) 8 = 6

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Question 03:

Solve: log_2 8 - log_(1/2) 8

###### Solution:
Given,
log_2 8 - log_(1/2) 8
we cannot use quotient rule here, since bases are not equal in each term
Hence,
we know that,
log_b a = (log a)/(log b)

⇒ log_2 (2^3) - (log 8)/(log (1/2))

But we know that, 1/n = n^-1

⇒ 3log_2 2 - (log 8)/(log 2^-1)

⇒ 3 - (log 8)/(-log 2)

⇒ 3 - (-log 8)/(log 2)

⇒ 3 + log_2 8
⇒ 3 + log_2 2^3
⇒ 3 + 3log_2 2
⇒ 3 + 3 * 1 = 6
∴ log_2 8 - log_(1/2) 8 = 6

Sorry, it's under construction !

Sorry, it's under construction !

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