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 !

Skills covered in the above question

`log_b a = 1/(log_a b)`


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

`log_b a = 1/(log_a b)`