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I found this manipulation of the equation for the error bound of the Bisection method here - http://www.maths.uniswa.sz/docs/m311/bisection.pdf

The error bound is $$\frac{b-a}{2^{n+1}} < \epsilon$$

They say "by taking logarithms we obtain" - $$n > \frac{\log(b-a) - \log(2\epsilon)}{\log (2)}$$

But that is not what I got when I maniuplated it. Here's what I did

$$\frac{b-a}{2^{n+1}} < \epsilon$$

$$2^{n+1}>\frac{b-a}{\epsilon}$$

$$2^{n}>\frac{b-a}{2\epsilon}$$

$$\log_2(2^{n})>\log_2(b-a) - \log_2(2\epsilon)$$

$$n>\log_2(b-a) - \log_2(2\epsilon)$$

So is this correct? Why does the answer from that pdf have a denominator of $\log(2)$?

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I would write the same thing as @Andre already mentioned. –  Seyhmus Güngören Sep 29 '12 at 19:55
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1 Answer

up vote 5 down vote accepted

You have not made a mistake.

You are using logarithms to the base $2$. The document uses logarithms to some unspecified other base, probably $e$, though it doesn't matter as long as one is consistent.

Whatever base $b$ the writer had in mind (it may have been $10$), we have $$\log_b(2^n)=n\log_b 2.$$ This accounts for the $\log 2$ in the denominator.

Your expression also "has" a log in the denominator, but it is $\log_2 2$, which is $1$.

The expression in the document has the advantage of being "calculator-ready."

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