For what values of $k$ does $(1+x)^{500+k}(1-x)^{500-k}$ exceed $10^9$?

We can compute the critical point for $(1+x)^{500+k}(1-x)^{500-k}$: $x=\frac{k}{500}$. The value at this point is $$ \left(1-\frac{k^2}{250000}\right)^{500}\left(\frac{500+k}{500-k}\right)^k $$ For $k\lt500$, this is an increasing function of $k$, the derivative of its log is $\log\left(\frac{500+k}{500-k}\right)$, and computing values shows that for $102\le k\lt500$, $(1+x)^{500+k}(1-x)^{500-k}$ will be bigger than $10^9$ for $x=\frac{k}{500}$.

For $k\lt102$, $(1+x)^{500+k}(1-x)^{500-k}$ will not exceed $10^9$ on $[0,1]$.

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We can approximate $$ \overbrace{\left(1-\frac{k^2}{250000}\right)^{500}}^{\sim e^{-\frac{k^2}{500}}}\overbrace{\left(\frac{500+k}{500-k}\right)^k\vphantom{\left(\frac{k^2}{2}\right)^5}}^{\sim e^{2\frac{k^2}{500}}} \sim e^{\frac{k^2}{500}} $$ If we solve $$ e^{\frac{k^2}{500}}=10^9 $$ we get $$ k=\sqrt{4500\log(10)}=101.792 $$