Question

question no 5 solve and send me fast

Answer 1

x = [a^(2n)-1]\[a^(2n+1],


x.a^(2n)+x=a^(2n)-1,


then

a^(2n)[x-1]=-1-x,


then
a^(2n)=(-1-x)/(x-1),


then

a^(2n)=(1+x)/(1-x),


taking log both sides, we have


2n.loga = log[(1+x)/(1-x)],


then


2n = log[(1+x)/(1-x)] ÷ loga,



therefore


n = 1/2 [log{(1+x)/(1-x)} ÷ loga]