How do you condense #4 ln 2 - ln 3 - 1#?

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Answer 1 · 2017-02-25T14:20:06

#4ln2-ln3-1=ln(16/(3e))#

#4ln2-ln3-1#

We know that: #alnb=lnb^a#, therefore,

#=ln2^4-ln3-1#
#=ln16-ln3-1#

We know that: #lna-lnb=ln(a*1/b)#, therefore,

#=ln(16*1/3)-1#
#=ln(16/3)-1#

We know that: #lne=1#, therefore,

#=ln(16/3)-lne#
#=ln(16/3*1/e)#
#=ln(16/(3e))#

Hence, #4ln2-ln3-1=ln(16/(3e))#