$\log_{16}(x y) = \log_{16}(x) + \log_{16}(y)=$

$\dfrac {\log_{4}(x)}{\log_4(16)} + \dfrac {\log_{4}(y)}{\log_4(16)}=$

$\dfrac {\log_4(x)}{\log_4(4^2)}+\dfrac {\log_4(y)}{\log_4(4^2)}=$

$\dfrac {\log_4(x)}{2}+\dfrac {\log_4(y)}{2}$

Results 1 to 9 of 9

- July 20th 2014, 05:46 AM #1

- Joined
- Jul 2014
- From
- uk
- Posts
- 6

## Please help me

Hello users,

I am trying to help my dad in working out a maths solution that he simply cannot understand. Maths had never been my strong subject but my dad is trying to reconnect with his past education questions done many years ago. I would be grateful if someone could provide a working solution as to how to calculate this question.

- July 20th 2014, 05:56 AM #2

- Joined
- Nov 2013
- From
- California
- Posts
- 4,121
- Thanks
- 1680

- July 20th 2014, 05:56 AM #3
## Re: Please help me

Can you use these logarithm laws: $\displaystyle \begin{align*} \log_{a}{ \left( n^p \right) } = p\log_a{(n)} \end{align*}$ and $\displaystyle \begin{align*} \log_{a}{\left( m\,n \right) } = \log_a{(m)} + \log_{a}{(n)} \end{align*}$ to simplify the right hand side to a single logarithm?

- July 20th 2014, 06:06 AM #4

- Joined
- Jul 2014
- From
- uk
- Posts
- 6

- July 20th 2014, 06:09 AM #5

- Joined
- Jul 2014
- From
- uk
- Posts
- 6

- July 20th 2014, 06:11 AM #6

- July 20th 2014, 06:57 AM #7

- Joined
- Jul 2014
- From
- uk
- Posts
- 6

## Re: Please help me

We have done the following we have done the first part by channing the bae from 16 to 4. Now to do the 2nd part we know from the first part that half of log x to the base 4 + 1/2 of log of y to the base 4 =3.5 (3 and half).

From the 2nd part of the equation is log of x to the base 4 - log of y to the base 4 = -8. This is gained by using the 2nd logarithm law. That is as far as we got, if my dad tries to solve these equations we are unable to do so due to some complications.

We don't know which law and how to use it to get the answer.

He done this type of questions in 1956 and his memory is not as strong as it used to be. It is more for nostalgic reasons he would like to reconnect to his maths.

- July 20th 2014, 06:59 AM #8

- Joined
- Jul 2014
- From
- uk
- Posts
- 6

## Re: Please help me

We have done the following we have done the first part by channing the bae from 16 to 4. Now to do the 2nd part we know from the first part that half of log x to the base 4 + 1/2 of log of y to the base 4 =3.5 (3 and half).

From the 2nd part of the equation is log of x to the base 4 - log of y to the base 4 = -8. This is gained by using the 2nd logarithm law. That is as far as we got, if my dad tries to solve these equations we are unable to do so due to some complications.

We don't know which law and how to use it to get the answer.

He done this type of questions in 1956 and his memory is not as strong as it used to be. It is more for nostalgic reasons he would like to reconnect to his maths.

- July 20th 2014, 10:00 AM #9

- Joined
- Dec 2011
- Posts
- 8

## Re: Please help me

Hey,

I think Dad has a bad memory. He wanted to get help on exactly the same questions (ie including your other post today) almost three years ago (Help with two log proof simultaneous questions please, thanks to @Siron for help back then). Thanks for posting this for him. I'll give him a ring and let him know. Can you mark this one 'Solved' when you next login please.

Dear MHF community: Our Dad's been very ill recently and he's thankfully started to get back into the swing of things, including picking up the old books he used to read. Appreciate everyone's help and kindness in feeding into the questions that just seem to be blockers for him.

Kindest regards.