Is the answer $\log\frac{2}{b^d+1}$?

Almost, but not quite =)

Then I think the answer is $\log2 - b^d$? Using the approximation $\log(1+x) \approx x$ when $x$ is small enough?

Oh wait, I parsed your first answer wrong. Your first answer is correct. It seems I'm having trouble reading tex today, sorry =S

Too much tex makes us tired. :)

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## 5 comments:

Is the answer $\log\frac{2}{b^d+1}$?

Almost, but not quite =)

Then I think the answer is $\log2 - b^d$? Using the approximation $\log(1+x) \approx x$ when $x$ is small enough?

Oh wait, I parsed your first answer wrong. Your first answer is correct. It seems I'm having trouble reading tex today, sorry =S

Too much tex makes us tired. :)

Post a Comment