Logarithm formula help

  • #1

Homework Statement



If log4 N=p and log12 N=q, show that

log3 48=
MSP310219e962cfa4gc2d1700001a0d133gdh9c3fh8.gif


Homework Equations





The Attempt at a Solution



I tried by substituting p and q into
MSP310219e962cfa4gc2d1700001a0d133gdh9c3fh8.gif
but i couldn't get the required answer. Can anyone help?
 

Answers and Replies

  • #2
Mentallic
Homework Helper
3,798
94


Use the formula [tex]\log_ab=\frac{\log_cb}{\log_ca}[/tex]
 
  • #3
VietDao29
Homework Helper
1,423
2


Homework Statement



If log4 N=p and log12 N=q, show that

log3 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 (a, x, y are all positive real number):

  • loga(xy) = logax + logay
  • [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: log4 N = p, and log12 N = q.

And we want to calculate: log348. If you look at that closely, you'll discover that: 48 = 4.12

So log348 = log3(4.12)

By playing with some of the identities I gave you above, you'll soon arrive at the desired answer. :)
 
  • #4


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?

log348
= log3 (4.12)
= log34 + log312
= (log3 N)/p + (log3N)/q
= (q log3N + p log3N)/pq
=[(p+q)log3N]/pq

Is it possible to continue?
 
  • #5
VietDao29
Homework Helper
1,423
2


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?

log348
= log3 (4.12)
= log34 + log312
= (log3 N)/p + (log3N)/q
= (q log3N + p log3N)/pq
=[(p+q)log3N]/pq

Is it possible to continue?

So, you are stuck in expressing log3N in terms of p = log4N, and q = log12N, 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? :)
 
  • #6


So, you are stuck in expressing log3N in terms of p = log4N, and q = log12N, 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
Science Advisor
Homework Helper
Gold Member
11,365
1,032


[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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