How do you find y given #dy/dx=2^(-x)# and #y = 1/(2log2# when x=1?

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Answer 1 · 2018-02-27T10:14:24

#y=1/ln2(1-2^(-x))#

assuming all the logs in this the question are to base #e#

as an introduction lets us do this

#y=2^(-x)#

#lny=ln2^(-x)#

#lny=-xln2#

differentiate wrt x

#1/y(dy)/(dx)=-ln2#

#(dy)/(dx)=-yln2=-2^(-x)ln2#

so comparing this with

#(dy)/(dx)=2^(-x)#

#y=int2^(-x)dx=-1/(ln2)2^(-x)+c#

#x=1, y=1/(2ln2)#

#1/(2ln2)=-1/ln2 2^(-1)+c#

#1/(2ln2)=-1/(2ln2)+c#

#c=2/(2ln2)=1/ln2#

#y=1/ln2(1-2^(-x))#