1. Couple of questions about logarithms and exponents

Which is the proper way of condensing this:

ln2-4ln3

Is it:

ln2+ln3^(-4)

or

ln2-ln3^4

Are either of these ways valid? And if so, which is the appropriate way? Or does that matter on the problem?

My other question.. Are these valid:
3^(x+1) = (3^x)*3
3^(x-1) = (3^x)/x or (3^x)(3^-1)

2. $ln2-4ln3=ln2-ln3^4=\frac{ln2}{ln3^4}$

$3^{x+1}=3^x3^1=3(3^x)$

$3^{x-1}=3^x3^{-1}=\frac{1}{3}(3^x)$

3. Originally Posted by Phire
Which is the proper way of condensing this:

ln2-4ln3

Is it:

ln2+ln3^(-4)

or

ln2-ln3^4

Are either of these ways valid? And if so, which is the appropriate way? Or does that matter on the problem?
ln2+ ln 3^(-4)= ln[2(3^-4)]= ln(2/81)
ln2- ln 3^4= ln[2/3^4]= ln(2/81)

They are exactly the same.

My other question.. Are these valid:
3^(x+1) = (3^x)*3
Yes, 3^(x+1)= 3^x*3^1= (3^x)*3

3^(x-1) = (3^x)/x or (3^x)(3^-1)
3^(x-1)= (3^x)(3^-1)= (3^x)/3. Is that what you meant instead of (3^x)/x?

4. Originally Posted by Gamma
$ln2-4ln3=ln2-ln3^4=\frac{ln2}{ln3^4}$

$3^{x+1}=3^x3^1=3(3^x)$

$3^{x-1}=3^x3^{-1}=\frac{1}{3}(3^x)$
$ln(a)-ln(b) \neq \frac{ln(a)}{ln(b)}$

Test with a = e and b=1

$ln(a)-ln(b) = ln(\frac{a}{b})$