# If a rough approximation for ln(5) is 1.609 how do you use this approximation and differentials to approximate ln(128/25)?

Mar 22, 2015

To approximate $\ln \left(\frac{128}{25}\right)$ using linear approximation and/or differentials we need a number near $\frac{128}{25}$ whose $\ln$ we know.

Clearly $\frac{128}{25}$ is somewhat near $\frac{125}{25}$

and $\frac{125}{25} = 5$ whose $\ln$ we were given.

The difference between $\ln \left(\frac{128}{25}\right)$ and $\ln \left(5\right)$ is approximately equal to the differential of $y = \ln x$

$\mathrm{dy} = \frac{1}{x} \mathrm{dx}$

To approximate near $5$, we use $\mathrm{dy} = \frac{1}{5} \mathrm{dx} = \frac{1}{5} \left(x - 5\right)$

With $x = \frac{128}{25}$, $x - 5 = \frac{3}{25} = \frac{12}{100} = 0.12$

and $\mathrm{dy} = \frac{1}{2} \left(0.12\right) = 0.024$

$\ln \left(\frac{128}{25}\right) = \ln \left(\frac{125}{25}\right) + \Delta y$

$\ln \left(\frac{128}{25}\right) \approx \ln \left(\frac{125}{25}\right) + \mathrm{dy} \approx 1.609 + 0.024 = 1.633$