How do you condense #4ln2#?

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Answer 1 · 2018-05-31T00:58:46

#ln16#

The key realization here is that we can use the logarithm property

#alnb=ln(b^a)#

Where the coefficient on #ln# becomes the exponent.

In our case, #a=4# and #b=2#. We can rewrite #4ln2# as

#ln(2^4)#

which can be further simplified as

#ln16#

Hope this helps!