MHB Find the value in terms of p and q

[mathlearn]

If $log_a 2 = P $ & $ log_a 3 = q$ find $log_a 72 $ in terms of $p$ & $q $

I'm not sure on how to begin this may be convert the logarithms to decimals

Many Thanks :)

 

[MarkFL]

We are given:

$$\log_a(2)=p$$ and $$\log_a(3)=q$$

Now, we may write:

$$\log_a(72)=\log_a\left(2^3\cdot3^2\right)$$

Can you now proceed to use the properties of logs to express this in terms of $p$ and $q$?

 

[mathlearn]

We are given:

$$\log_a(2)=p$$ and $$\log_a(3)=q$$

Now, we may write:

$$\log_a(72)=\log_a\left(2^3\cdot3^2\right)$$

Can you now proceed to use the properties of logs to express this in terms of $p$ and $q$?

Thank you very much MarkFL (Happy) (Party)

Then expressing this in terms of $p$ & $q$,

It would be $p=3$ & $q=2$

May I know how the power of 3 on 2 & the power of 2 on 3 get derived below in the RHS

$\displaystyle \log_a(72)=\log_a\left(2^3\cdot3^2\right)$

 

[MarkFL]

Thank you very much MarkFL (Happy) (Party)

Then expressing this in terms of $p$ & $q$,

It would be $p=3$ & $q=2$

May I know how the power of 3 on 2 & the power of 2 on 3 get derived below in the RHS

$\displaystyle \log_a(72)=\log_a\left(2^3\cdot3^2\right)$

No, $p$ and $q$ don't somehow change values here. We observe that by prime factorization we have:

$$72=8\cdot9=2^3\cdot3^2$$

Hence:

$$\log_a(72)=\log_a\left(2^3\cdot3^2\right)$$

Now, you want to use the properties of logs, specifically:

$$\log_a(b^c)=c\cdot\log_a(b)$$

$$\log_a(bc)=\log_a(b)+\log_a(c)$$

to further simplify. What do you get?

 

[mathlearn]

No, $p$ and $q$ don't somehow change values here. We observe that by prime factorization we have:

$$72=8\cdot9=2^3\cdot3^2$$

Hence:

$$\log_a(72)=\log_a\left(2^3\cdot3^2\right)$$

Now, you want to use the properties of logs, specifically:

$$\log_a(b^c)=c\cdot\log_a(b)$$

$$\log_a(bc)=\log_a(b)+\log_a(c)$$

to further simplify. What do you get?

Thank you very much MarkFL (Happy) (Party)

My apologies! for being a little late in replying,

Using the given properties of logarithms,

$\log_a72 = \log_a(2^3 \cdot 3^2)\\
=\log_a (2^3) + \log_a (3^2)\\
= 3 \log_a2 + 2\log_a3\\
= 3p + 2q$

Correct ? (Smile)

 

[MarkFL]

Yes, that looks good. :D