### Logarithms.

Another rule:
log_b a = 1/(log_a b)

Subjects > Mathematics > Logarithms

Question 09:

Without using logaithmic tables or calculator, evaluate:
(log (1/4) + log 64)/(log 32 - log (1/8))

###### Solution:
Given,
(log (1/4) + log 64)/(log 32 - log (1/8))

both terms are common logarithms, hence we can use product rule and quotient rule
i.e
⇒ (log (1/4 * 64))/(log ((32)/(1/8))

⇒ (log 16)/(log 256) = log_256 16

but,
log_b a = 1/(log_a b)

Hence,
log_256 16 = 1/(log_16 256)

⇒ log_256 16 = 1/(log_16 16^2)

⇒ log_256 16 = 1/(2log_16 16) = 1/2

∴ (log (1/4) + log 64)/(log 32 - log (1/8)) = 1/2

Sorry, it's under construction !

Sorry, it's under construction !

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Related Lessons

Question 09:

Without using logaithmic tables or calculator, evaluate:
(log (1/4) + log 64)/(log 32 - log (1/8))

###### Solution:
Given,
(log (1/4) + log 64)/(log 32 - log (1/8))

both terms are common logarithms, hence we can use product rule and quotient rule
i.e
⇒ (log (1/4 * 64))/(log ((32)/(1/8))

⇒ (log 16)/(log 256) = log_256 16

but,
log_b a = 1/(log_a b)

Hence,
log_256 16 = 1/(log_16 256)

⇒ log_256 16 = 1/(log_16 16^2)

⇒ log_256 16 = 1/(2log_16 16) = 1/2

∴ (log (1/4) + log 64)/(log 32 - log (1/8)) = 1/2

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

Related Lessons