SOLUTION 18: Begin with the function
$$f(x)= \log(100-x)$$
and choose
$$x-values: 0 \rightarrow h^4$$
so that
$$\Delta x = h^4 - 0 = h^4$$

$\Big($Recall that $D\{ \log(g(x)) \} = \displaystyle{ 1 \over g(x) } \cdot g'(x) \cdot { 1 \over \ln(10) }. \Big)$

The derivative of $\ y=f(x) \$ is
$$f'(x)= \displaystyle{ 1 \over 100-x } \cdot (-1) \cdot { 1 \over \ln(10) }$$ $$= \displaystyle{ -1 \over (-1)(x-100) } \cdot { 1 \over \ln(10) }$$ $$= \displaystyle{ 1 \over x-100 } \cdot { 1 \over \ln(10) }$$
The exact change of $y-$values is
$$\Delta y = f(h^4) - f(0)$$ $$= \log(100-(h^4)) - \log(100+(0))$$ $$= \log(100-h^4) - \log(100)$$

$\Big($ Recall that $\log(10^n)=n .\Big)$

$$= \log(100-h^4) - \log(10^2)$$ $$= \log(100-h^4) - 2$$
The Differential is
$$dy = f'(0) \ \Delta x$$ $$= \displaystyle{ 1 \over ((0)^4) - 100 } \cdot { 1 \over \ln(10) } \cdot (h^4)$$ $$= \displaystyle{ 1 \over -100 } \cdot { 1 \over \ln(10) } \cdot (h^4)$$ $$= \displaystyle{ -1 \over 100 \ln(10) } h^4$$
Since $h$ is "small" we will assume that
$$\Delta y \approx dy \ \ \ \ \longrightarrow$$ $$\log(100-h^4) - 2 \approx \displaystyle{ -1 \over 100 \ln(10) } h^4 \ \ \ \ \longrightarrow$$ $$\log(100-h^4) \approx 2 - \displaystyle{ 1 \over 100 \ln(10) } h^4$$