Logarithms.

Remember this:
`log_b a = (log a)/(log b)`

Subjects > Mathematics > Logarithms


Question 08:

Solve for `x`, given that: `log_2 5x - log_4 5x = 3`

Solution:
Given,
`log_2 5x - log_4 5x = 3`
we cannot use quotient rule now, since the terms are of different bases
Hence,
from: `log_b a = (log a)/(log b)`

⇒ `log_2 5x - (log 5x)/(log 2^2) = 3`

⇒ `log_2 5x - (log 5x)/(2log 2) = 3`

⇒ `log_2 5x - 1/2 * (log 5x)/(log 2) = 3`

⇒ `log_2 5x - 1/2 * log_2 5x = 3`
let `log_2 5x = a`
Now,
⇒ `a - a/2 = 3`
⇒ `a/2 = 3` ⇒ `a = 2` x 3 = 6
Hence,
`log_2 5x = a = 6`
⇒ `5x = 2^6`
`5x = 64` ⇒ `(5x)/5 = (64)/5 = 12.8`
∴ The value of `x` is `12.8`

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

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


Question 08:

Solve for `x`, given that: `log_2 5x - log_4 5x = 3`

Solution:
Given,
`log_2 5x - log_4 5x = 3`
we cannot use quotient rule now, since the terms are of different bases
Hence,
from: `log_b a = (log a)/(log b)`

⇒ `log_2 5x - (log 5x)/(log 2^2) = 3`

⇒ `log_2 5x - (log 5x)/(2log 2) = 3`

⇒ `log_2 5x - 1/2 * (log 5x)/(log 2) = 3`

⇒ `log_2 5x - 1/2 * log_2 5x = 3`
let `log_2 5x = a`
Now,
⇒ `a - a/2 = 3`
⇒ `a/2 = 3` ⇒ `a = 2` x 3 = 6
Hence,
`log_2 5x = a = 6`
⇒ `5x = 2^6`
`5x = 64` ⇒ `(5x)/5 = (64)/5 = 12.8`
∴ The value of `x` is `12.8`

Sorry, it's under construction !

Sorry, it's under construction !

Skills covered in the above question

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