How do you solve #8(4^(6-2x))+13=41#?

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Answer 1 · 2017-07-19T14:16:07

#x~=2.548#

#8(4^(6-2x))+13=41#

#=>8(4^(6-2x))=41-13=28#

or #4^(6-2x)=28/8=7/2#

or #2^(12-4x)=7/2=3.5#

Therefore #12-4x=log_2 3.5=log3.5/log2#

or #12-4x=0.54407/0.30103=1.807355#

i.e. #4x=12-1.807355=10.192645#

and #x=10.192645/4~=2.548#