The value of `log_8 32 - log_n sqrt n` has to be determined

`log_8 32 - log_n sqrt n`

=> `(log_2 32)/(log_2 8) - log_n n^(1/2)`

=> `(log_2 2^5)/(log_2 2^3) - log_n n^(1/2)`

=>` (5*log_2 2)/(3*log_2 2) - (1/2)*log_n n`

=> `5/3 - 1/2`

=> `7/6`

**The required value of `log_8 32 - log_n sqrt n` = **`7/6`

You need to evaluate `log_8 32 - log _ n sqrt n` , hence you shoud write 32 as product `8*4` and you need to write `sqrt n = n^(1/2)` such that:

`log_8 32 - log _ n sqrt n = log_8(8*4) - log _ n (n^(1/2))`

You should use the properties of logarithms, hence you should convert the logarithm of product into sum of logarithms and you need to use power rule such that:

`log_8 32 - log _ n sqrt n = log_8 8 + log_84 - (1/2)log _ n n`

`log_8 32 - log _ n sqrt n = 1 + 1/(log_4 8) - 1/2`

`log_8 32 - log _ n sqrt n = 1/2 + 1/(log_4 4 + log_4 2)`

`log_8 32 - log _ n sqrt n = 1/2 + 1/(1 + 1/(log_2 4))`

`log_8 32 - log _ n sqrt n = 1/2 + 1/(1 + 1/2)`

`log_8 32 - log _ n sqrt n = 1/2 + 2/3`

You need to bring the fraction to the right to a common denominator such that:

`log_8 32 - log _ n sqrt n = (3+4)/6`

`log_8 32 - log _ n sqrt n = 7/6`

**Hence, evaluating the difference `log_8 32 - log _ n sqrt n=> log_8 32 - log _ n sqrt n = 7/6` .**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now