Question

Consider functions f and g.

Answer 1

It depends on what method you're using to approximate. Newton's method is a popular choice. Let $h(x) = g(x) - f(x) = 3\log_{10}(x-2) - \log_{10}(x)$, with derivative

$h'(x) = \dfrac1{\ln(10)} \left(\dfrac{3}{x-2} - \dfrac1x\right)$

which follows from

$\dfrac{d}{dx} \log_{10}(x) = \dfrac{d}{dx} \dfrac{\ln(x)}{\ln(10)} = \dfrac1{\ln(10)\,x}$

Judging by the plots, we have $f(x)=g(x)$ for some $x$ between 3 and 4. So let's take an initial approximation of $x_0 = 3$. Newton's method involves taking the tangent line approximation to $f(x)$ at $x=x_0$, then using the $x$-intercept of this tangent line as the next approximation, $x_1$.

The linear approximation at $x_0=3$ is

$h(x) \approx h(x_0) + h'(x_0) (x - x_0) \approx 1.15812x - 3.95148$

with intercept at $x_1 \approx 3.41198$.

Repeat until you reach the desired threshold of accuracy. The next linear approximation at $x_1$ is

$h(x) \approx h(x_1) + h'(x_1) (x - x_1) \approx 0.79545 x-2.79758$

with intercept $x_2 \approx 3.51698$.

The next approximation at $x_2$ is

$h(x) \approx h(x_2) + h'(x_2) (x - x_2) \approx 0.735382 x-2.58956$

with intercept $x_3 \approx 3.52137$.

And so on. The actual solution has an approximate value of about

3.521379706804567569604081

so after just 3 approximations we're already less than $10^{-5}$ away.