Express as a single logarithm

I dont understand this math problem can someone help? Lots of points to the best answer.

(log(a,m) - log(a,n)) +3log(a,k)

3/4ln16-ln(4^2-3^2-2)

For the first one, I am going to assume that 'a' is the base. When you have a same base, you can combine logs as such (I am going to use ln so you know that I have the same base which I dont have to write out): lnX + lnY = ln(XY) and lnX - lnY = ln(X/Y)

1.) (log(base a)m - log(base a)n) + 3log(base a)k
log(base a)(m/n) + log(base a)(k^3)..................(you can move the coefficient as a power of 'k')
log(base a)(mk^3/n)....and that is your final answer.

2.) Lets start of by cleaning some mess up. Move the 3/4 as a power of 16 (like we did last time). The new number should be 8. Then, simplify 4^2-3^2-2. You should end up with 5. Now, write the simpler logorithim expression: ln8-ln5. Recall how to combine logs with the same base. (look at what I wrote earlier). ln8 - ln5 = ln(8/5) which is approxmately 0.470003629.

(1) What is "(log(a,m) - log(a,n)) + 3log(a,k)" supposed to mean?
Is "," is supposed be "^"?

If so, log(a^m) - log(a^n)) + 3 log(a^k)
= log(a^m / a^n) + log((a^k)³)
= log(a^m (a^k)³ / a^n)



(2)

3/4ln16-ln(4^2-3^2-2)
= 3/4 ln(16) - ln(4^2 - 3^2 -2)
= ln(16^(3/4)) - ln(16 - 9 - 2)
= ln(8) - ln(5)
= ln(8/5)



Memorize the following log identities:

log(xy) = log(x) + log(y)
log(x^y) y log(x)

We are going to use log(a,m) - log(a,n) = log(a,m/n) and
n*log(a,m) = log(a, m^n)
1. = log(a,m/n) + log(a,k^3)
= log(a, mk^3/n)

2. ln(16^(3/4)) - ln(16-9-2) = ln((2^4)^(3/4) - ln(5)
= ln(2^(4*(3/4)) - ln(5) = ln(2^3)-ln(5) = ln(8/5)