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The correct answers are discussed in later posts. I haven't updated my first posted for a while.Check your calculation. It should be(\ln a = \frac{1}{2} \ln (2 + a))(a^2=2+a \Rightarrow a =2)There are easier way to do this without using log but the idea is the same.
The correct answers are discussed in later posts. I haven't updated my first posted for a while.
Check your calculation. It should be
(\ln a = \frac{1}{2} \ln (2 + a))
(a^2=2+a \Rightarrow a =2)
There are easier way to do this without using log but the idea is the same.
