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.