How do you solve #4^ { y + 9} = 11#?

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Answer 1 · 2017-06-09T07:55:39

#y=-7.27" "(2dp)#

take logs both sides

#4^(y+9)=11#

#ln[4^(y+9)]=ln11#

#(y+9)ln4=ln11#

#y+9=(ln11)/(ln4)#

#y=(ln11)/(ln4)-9#

#y= -7.270284191#

#y=-7.27" "(2dp)#