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 $x=10.192645/4~=2.548$

How do you solve 8(4^(6-2x))+13=418(4^(6-2x))+13=41?