It depends on what method you're using to approximate. Newton's method is a popular choice. Let [tex]h(x) = g(x) - f(x) = 3\log_{10}(x-2) - \log_{10}(x)[/tex], with derivative
[tex]h'(x) = \dfrac1{\ln(10)} \left(\dfrac{3}{x-2} - \dfrac1x\right)[/tex]
which follows from
[tex]\dfrac{d}{dx} \log_{10}(x) = \dfrac{d}{dx} \dfrac{\ln(x)}{\ln(10)} = \dfrac1{\ln(10)\,x}[/tex]
Judging by the plots, we have [tex]f(x)=g(x)[/tex] for some [tex]x[/tex] between 3 and 4. So let's take an initial approximation of [tex]x_0 = 3[/tex]. Newton's method involves taking the tangent line approximation to [tex]f(x)[/tex] at [tex]x=x_0[/tex], then using the [tex]x[/tex]-intercept of this tangent line as the next approximation, [tex]x_1[/tex].
The linear approximation at [tex]x_0=3[/tex] is
[tex]h(x) \approx h(x_0) + h'(x_0) (x - x_0) \approx 1.15812x - 3.95148[/tex]
with intercept at [tex]x_1 \approx 3.41198[/tex].
Repeat until you reach the desired threshold of accuracy. The next linear approximation at [tex]x_1[/tex] is
[tex]h(x) \approx h(x_1) + h'(x_1) (x - x_1) \approx 0.79545 x-2.79758[/tex]
with intercept [tex]x_2 \approx 3.51698[/tex].
The next approximation at [tex]x_2[/tex] is
[tex]h(x) \approx h(x_2) + h'(x_2) (x - x_2) \approx 0.735382 x-2.58956[/tex]
with intercept [tex]x_3 \approx 3.52137[/tex].
And so on. The actual solution has an approximate value of about
3.521379706804567569604081
so after just 3 approximations we're already less than [tex]10^{-5}[/tex] away.
