Logarithms.

is the inverse function to exponentation, as it answers the question, how many of one number do we multiply to get another number.
If `log_b a = x` then `a = b^x`

Subjects > Mathematics > Logarithms


Question 05:

Find the value of `log root(3)(1/27)`

Solution:
Given,
`log root(3)(1/27)`
But we know that
`root(n)(x) = x^(1/n)`

Hence,
⇒ `log (1/27)^(1/3)`
but `27 = 3^3`
⇒ `log (1/(3^3))^(1/3)`

But also,
`1/(3^3) = (1/3)^3`
Hence,
⇒ `log (1/3)^(3 * 1/3)` = `log (1/3)`
⇒ `log (1/3) = log 3^-1 = -log 3`
From mathematical table,
log 3 `~~` 0.4771
Hence,
⇒ `-log 3 = -0.4771`
∴ `log root(3)(1/27) ~~ -0.4771`

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

power rule of logarithms


Question 05:

Find the value of `log root(3)(1/27)`

Solution:
Given,
`log root(3)(1/27)`
But we know that
`root(n)(x) = x^(1/n)`

Hence,
⇒ `log (1/27)^(1/3)`
but `27 = 3^3`
⇒ `log (1/(3^3))^(1/3)`

But also,
`1/(3^3) = (1/3)^3`
Hence,
⇒ `log (1/3)^(3 * 1/3)` = `log (1/3)`
⇒ `log (1/3) = log 3^-1 = -log 3`
From mathematical table,
log 3 `~~` 0.4771
Hence,
⇒ `-log 3 = -0.4771`
∴ `log root(3)(1/27) ~~ -0.4771`

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

power rule of logarithms