Logarithms problem

[jonnburton]

Hi everyone,

Could anybody tell me how to write this as a single logarithm:

\(\displaystyle 2log_a 6 - log_a 4\)

I've seen the answer should be \(\displaystyle log_a9\) according to the book but I'm not sure how to get this answer. The fact that the log has a coefficient is throwing me and I'm not sure how to deal with that coefficient.

This is what I did:

\(\displaystyle log_ax - log_ay = log_a (\frac{x}{y})\)

\(\displaystyle log_a (\frac{12}{4})\)

\(\displaystyle log_a3\)

Really, I would like to know how to deal with the coefficient 6, because that's what's confusing me here...

Thanks in advance,

 

[jonnburton]

Hi everyone,

Could anybody tell me how to write this as a single logarithm:

\(\displaystyle 2log_a 6 - log_a 4\)

I've seen the answer should be \(\displaystyle log_a9\) according to the book but I'm not sure how to get this answer. The fact that the log has a coefficient is throwing me and I'm not sure how to deal with that coefficient.

This is what I did:

\(\displaystyle log_ax - log_ay = log_a (\frac{x}{y})\)

\(\displaystyle log_a (\frac{12}{4})\)

\(\displaystyle log_a3\)

Really, I would like to know how to deal with the coefficient 6, because that's what's confusing me here...

Thanks in advance,

OH, I've just looked through it all again and recalled the other law of logarithms - that \(\displaystyle K log_a x = log_a(x^k)\)... It should be possible for me to figure it out now!

Sometimes there is just too much new information for me to absorb it all and know when to apply it. This is a case in point!

 

[DrPhil]

OH, I've just looked through it all again and recalled the other law of logarithms - that \(\displaystyle k log_a x = log_a(x^k)\)... It should be possible for me to figure it out now!

Sometimes there is just too much new information for me to absorb it all and know when to apply it. This is a case in point!

And sometimes, just writing out the problem for somebody else to read helps your own understanding!

Do what you did before, but with 6^2 instead of 2×6

 

[jonnburton]

Yes, that's true DrPhil; it wouldn't be the first time I have prepared a post on something, only for it to 'click' while doing so!

Yep, I got the correct answer doing it this way. It led to

\(\displaystyle log_a (\frac{36}{4}) = log_a 9\)