Logarithm Formula Help: Solving for Log3 48 with Given Log4 N and Log12 N

[Michael_Light]

Homework Statement

If log₄ N=p and log₁₂ N=q, show that

log₃ 48=

Homework Equations

The Attempt at a Solution

I tried by substituting p and q into

but i couldn't get the required answer. Can anyone help?

 

[Mentallic]

Use the formula $$\log_ab=\frac{\log_cb}{\log_ca}$$

 

[VietDao29]

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):

• log_a(xy) = log_ax + log_ay
• $$\log_{a} \left( \frac{x}{y} \right) = \log_{a}(x) - \log_{a}(y)$$
• $$\log_{a} x = \frac{1}{\log_{x} a}$$
• $$\log_{a} b \times \log_{b} c = \log_{a} c$$ or $$\log_{a} b =\frac{\log_{c} b}{\log_{c} a}$$

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. :)

 

[Michael_Light]

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?

 

[VietDao29]

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:

$$\log_{3} N = \frac{1}{\log_{N} 3} = \frac{1}{\log_{N} \left( \frac{12}{4} \right)} = ...$$

Can you finish it from here? :)

 

[Michael_Light]

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:

$$\log_{3} N = \frac{1}{\log_{N} 3} = \frac{1}{\log_{N} \left( \frac{12}{4} \right)} = ...$$

Can you finish it from here? :)

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

 

[SammyS]

$$\log_{4}\,N=p\quad\Leftrightarrow\quad4^{p}=N\quad\text{ and }\quad\log_{12}\,N=q\quad\Leftrightarrow\quad12^{q}=N$$