Help with logarithms

Question

darkbluechocobo · Answer

4[logm + log(m+5)] − 3log(m + 5)

darkbluechocobo · Answer

i need to write this as one logarithm

e.mccormick · Answer

Know your basic log rules?

$\log_b(xy)=\log_bx+log_by$

$\log_b\left(\dfrac xy\right)=\log_bx-log_by$

$\log_b(x^y)=y\log_bx$

darkbluechocobo · Answer

yes just a bit lost x.x

e.mccormick · Answer

OK, so look at these one at a time.  You have [ ], so you have a clear order of operations to start with.

darkbluechocobo · Answer

alright one moment then

darkbluechocobo · Answer

so the first bracket would be 4[log(m^2+5m)]-3log(m+5

e.mccormick · Answer

Well, might not want to distribute yet... the subtracted one looks like something that might cancel... but still, it is the correct direction.

Then you have the numbers out fron that become powers.

darkbluechocobo · Answer

$$\log \frac{ m }{ (m+5)^2 }$$

darkbluechocobo · Answer

well $$4(\log \frac{ m }{ (m+5)^2 }$$

e.mccormick · Answer

Hmmm....  I think you missed a few things in there.  You jad a 4 out fron of the first one.  That power needs to be handled before the fraction, just like the 3 out front of the last one did.

darkbluechocobo · Answer

sorry back

darkbluechocobo · Answer

So I distribute the 4 ?

e.mccormick · Answer

This $4[\log m + \log(m+5)] - 3\log(m + 5) \ $ to
this $4 \log(m(m+5)) - 3\log(m + 5) $ was fine.

The next step, with the powers, what do you get for just putting the powers in the proper place.  Use ( ) grouping inside the ( ) for the log.

e.mccormick · Answer

Well, I had the whole thing worked out... here is the next step.  Hope you can see how it could go to a fraction from there, which would allow distribution of an exponent and cancelation.

$ \log((m(m+5))^4) - \log((m + 5)^3) $