### 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 !

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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

Related Lessons