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Find all solutions of $$5 x+\ln x=10000$$, correct to 4 decimal places; use the Newton Method.
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Let $$f(x)=5 x+\ln x-10000$$. We need to approximate the root(s) of the equation $$f(x)=0$$. The function $$f$$ is only defined for positive $$x$$. Note that the function is steadily increasing, since $$f^{\prime}(x)=5+1 / x>0$$ for all positive $$x$$. It follows that the function can be 0 for at most one value of $$x$$. It is easy to verify that $$f(1)<0$$ and $$f(2000)>0$$, and therefore the equation has a root in the interval $$(1,2000)$$.

The Newton Method iteration is easy to set up. We get
$x_{n+1}=x_{n}-\frac{5 x_{n}+\ln x_{n}-10000}{5+1 / x_{n}} .$
We could simplify the right hand side somewhat. This is probably not worthwhile.

Now we need to choose $$x_{0}$$. The idea is that even when $$x$$ is large, $$\ln x$$ is by comparison quite small. So as a first approximation we can forget about the $$\ln x$$ term, and decide that $$f(x)$$ is approximately $$5 x-10000$$. Thus the root of our original equation must be near $$x=2000$$.

Shall we choose $$x_{0}=2000$$ ? It is sensible to do so. But we can do better. Note that $$\ln (2000)$$ is about 7.6. So we can take $$5 x_{0} \approx$$ $$10000-7.6$$. Let $$x_{0}=1998.48$$.

A quick computation gives $$x_{1}=1998.479972$$. This agrees with $$x_{0}$$ to 4 decimal places, so the answer, correct to 4 decimal places, should be 1998.4800. If we feel like it, we can show by the usual "sign change" procedure that this answer is indeed correct to 4 places.

Note. If we start with $$x_{0}=2000$$, it turns out that $$x_{1}=1998.479972$$, so perhaps the extra thinking that went into starting with $$1998.48$$ was unnecessary. But it illustrates the fact that in some cases we can get an extremely accurate estimate of a root without bringing out heavy machinery.
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