- #1

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## Homework Statement

If log

_{4}N=p and log

_{12}N=q, show that

log

_{3}48=

## Homework Equations

## The Attempt at a Solution

I tried by substituting p and q into

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- Thread starter Michael_Light
- Start date

- #1

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If log

log

I tried by substituting p and q into

- #2

Mentallic

Homework Helper

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Use the formula [tex]\log_ab=\frac{\log_cb}{\log_ca}[/tex]

- #3

VietDao29

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## Homework Statement

If log_{4}N=p and log_{12}N=q, show that

log_{3}48= View attachment 32226

## Homework Equations

## The Attempt at a Solution

I tried by substituting p and q into View attachment 32226 but i couldn't get the required answer. Can anyone help?

Well, can you post your work, so that we can help you continue it? You are on the right track. Some of the logarithmic identities you should remember is (

- log
_{a}(*xy*) = log+ log_{a}x_{a}y - [tex]\log_{a} \left( \frac{x}{y} \right) = \log_{a}(x) - \log_{a}(y)[/tex]
- [tex]\log_{a} x = \frac{1}{\log_{x} a}[/tex]
- [tex]\log_{a} b \times \log_{b} c = \log_{a} c[/tex] or [tex]\log_{a} b =\frac{\log_{c} b}{\log_{c} a}[/tex]

There's another way.

Note that, we have: log

And we want to calculate: log

So log

By playing with some of the identities I gave you above, you'll soon arrive at the desired answer. :)

- #4

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Thanks! I finally managed to solve the question by proving that (p+q)/(p-q) is equal to L.H.S.

Just to ask is it possible for me to obtain (p+q)/(p-q) from what i have done below? How should i continue?

log

= log

= log

= (log

= (q log

=[(p+q)log

Is it possible to continue?

- #5

VietDao29

Homework Helper

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Just to ask is it possible for me to obtain (p+q)/(p-q) from what i have done below? How should i continue?

log_{3}48

= log_{3}(4.12)

= log_{3}4 + log_{3}12

= (log_{3}N)/p + (log_{3}N)/q

= (q log_{3}N + p log_{3}N)/pq

=[(p+q)log_{3}N]/pq

Is it possible to continue?

So, you are stuck in expressing log

[tex]\log_{3} N = \frac{1}{\log_{N} 3} = \frac{1}{\log_{N} \left( \frac{12}{4} \right)} = ...[/tex]

Can you finish it from here? :)

- #6

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So, you are stuck in expressing log_{3}Nin terms ofp= log_{4}N, andq= log_{12}N, right? First, you should notice that 3 = 12/4. So, we have:

[tex]\log_{3} N = \frac{1}{\log_{N} 3} = \frac{1}{\log_{N} \left( \frac{12}{4} \right)} = ...[/tex]

Can you finish it from here? :)

Thousands of thanks VietDao29! I solved it. Thanks again. ^^

- #7

SammyS

Staff Emeritus

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Gold Member

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[tex]\log_{4}\,N=p\quad\Leftrightarrow\quad4^{p}=N\quad\text{ and }\quad\log_{12}\,N=q\quad\Leftrightarrow\quad12^{q}=N[/tex]

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