Solve the following equation...? #2^(4x) - 5(2^(2x - 1/2)) + 2 = 0#

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Answer 1 · 2017-11-26T21:01:12

#x=ln((25+-sqrt(609))/(2sqrt(2)))/(ln4)#

#2^(4x)-5(2^(2x-1/2))+2=0<=>#

#2^((2x)^2)-5*2^(2x)color(red)(xx)5*2^(-1/2)+2=0<=>#

#(2^(2x))^2-(25/sqrt(2))2^(2x)+2=0<=>#

Now the quadratic equation should be easy to see.
You have to replace #2^(2x)# with an y.

#<=> y^2−(25/(√2))y+2=0#

#y=(25/sqrt(2)+-sqrt(625/2-2*2*2))/2#

#y=(25/sqrt(2)+-sqrt(609/2))/2#

#2^(2x)=y=(25/sqrt(2)+-sqrt(609/2))/2#

Appyling logarithms:

#2xln2=ln((25+-sqrt(609))/(2sqrt(2)))#

#x=ln((25+-sqrt(609))/(2sqrt(2)))/(2ln2)#

#x=ln((25+-sqrt(609))/(2sqrt(2)))/(ln4)#