How do you solve #9^(n+10)+3=81#?

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Answer 1 · 2016-12-18T17:40:21

#n=-7/2#

#9^(n+10)+3=81#
#3^(2n+10)+3^1=3^4#
#2n+10+1=4#
#2n=4-11#
#2n=-7#
#n=-7/2#
substitute# 2n=-7#
#-7+10+1=4#
#4=4#

Answer 2 · 2016-12-23T21:13:17

#n ~~ -8.017#

#9^(n+10) +3 = 81#

#9^(n+10) = 78" "# the variable is in the index, so use logs

#log9^(n+10) = log78" "larr#take logs of both sides

#(n+10)log9 = log78" "larr# log law #loga^b = bloga#

#n+10= log78/log9#

#n = log78/log9 -10" "larr# use a calculator at this point

#n = -8.017176#