Write the following expression as a single logarithm: 4ln2x + ln(6/x) - 2ln2x

Question

So I did...
4ln2x + ln(6/x) - 2ln2x
= ln2x^4 + ln(6/x) - ln2x^2
= ln16x^4 + ln(6/x) - ln4x^2
= ln16x^(2)(2) + ln(6/x) - ln4x^2
= 2ln16x^2 + ln6 - lnx - ln4x^2

If my calculations so far are correct, then what do I do next in order to express as a single logarithm?

Answer 1

You could use some parentheses.
I'd have done it like this, combining the ln(2x) terms first:

4ln(2x) + ln(6/x) - 2ln(2x)
2ln(2x) + ln(6/x)
2ln2 + 2lnx + ln6 - lnx
ln4 + ln6 + lnx
ln(24x)

Or, doing it kind of your way,

4ln2x + ln(6/x) - 2ln2x
ln(16x^4) + ln(6/x) - ln(4x^2)
ln[(16x^4*6)/(4x^3)]
ln(24x)