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 02:

Evaluate: `log_49 7 + log_8 64`

Solution:
Given,
`log_49 7 + log_8 64`

we cannot use the product rule of logarithms since the terms have different bases
Now,
⇒ `log_49 7 + log_8 (8^2)`
⇒ `log_49 7 + 2log_8 8`

But, since `log_a a = 1`
Hence, `log_8 8 = 1`

⇒ `log_49 7 + 2log_8 8` = `log_49 7 + 2`

But, `log_b a = (log a)/(log b)`
Hence,
⇒ `(log 7)/(log 49) + 2`

⇒ `(log 7)/(log 7^2) + 2`

⇒ `(log 7)/(2log 7) + 2`

⇒ `1/2 * (log 7)/(log 7) + 2`

⇒ `1/2 * log_7 7 + 2`
Again, `log_7 7 = 1`
Hence,
⇒ `1/2 * 1 + 2 = 2 1/2 = 2.5`

∴ `log_49 7 + log_8 64` = 2.5

Sorry, it's under construction !

Skills covered in the above question

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


Question 02:

Evaluate: `log_49 7 + log_8 64`

Solution:
Given,
`log_49 7 + log_8 64`

we cannot use the product rule of logarithms since the terms have different bases
Now,
⇒ `log_49 7 + log_8 (8^2)`
⇒ `log_49 7 + 2log_8 8`

But, since `log_a a = 1`
Hence, `log_8 8 = 1`

⇒ `log_49 7 + 2log_8 8` = `log_49 7 + 2`

But, `log_b a = (log a)/(log b)`
Hence,
⇒ `(log 7)/(log 49) + 2`

⇒ `(log 7)/(log 7^2) + 2`

⇒ `(log 7)/(2log 7) + 2`

⇒ `1/2 * (log 7)/(log 7) + 2`

⇒ `1/2 * log_7 7 + 2`
Again, `log_7 7 = 1`
Hence,
⇒ `1/2 * 1 + 2 = 2 1/2 = 2.5`

∴ `log_49 7 + log_8 64` = 2.5

Sorry, it's under construction !

Skills covered in the above question

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

Power rule of logarithms