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`

Skills covered in the above question

`log_b a = (log a)/(log b)` and Power rule of 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`

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

`log_b a = (log a)/(log b)`

Power rule of logarithms