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

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Answer 1 · 2017-07-19T14:16:07

$x ≅ 2.548$ $x~=2.548$

$8 (4^{6 - 2 x}) + 13 = 41$ $8(4^(6-2x))+13=41$

$\Rightarrow 8 (4^{6 - 2 x}) = 41 - 13 = 28$ $=>8(4^(6-2x))=41-13=28$

or $4^{6 - 2 x} = \frac{28}{8} = \frac{7}{2}$ $4^(6-2x)=28/8=7/2$

or $2^{12 - 4 x} = \frac{7}{2} = 3.5$ $2^(12-4x)=7/2=3.5$

Therefore $12 - 4 x = {log}_{2} 3.5 = \frac{log 3.5}{log 2}$ $12-4x=log_2 3.5=log3.5/log2$

or $12 - 4 x = \frac{0.54407}{0.30103} = 1.807355$ $12-4x=0.54407/0.30103=1.807355$

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

and $x = \frac{10.192645}{4} ≅ 2.548$ $x=10.192645/4~=2.548$

How do you solve 8(4^(6-2x))+13=418(46−2x)+13=418(4^(6-2x))+13=41?