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

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

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Answer 1 · 2015-03-22T14:35:41

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

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

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

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

$d y = \frac{1}{x} d x$ $dy=1/x dx$

To approximate near $5$ $5$ , we use $d y = \frac{1}{5} d x = \frac{1}{5} (x - 5)$ $dy = 1/5 dx= 1/5 (x-5)$

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

and $d y = \frac{1}{2} (0.12) = 0.024$ $dy=1/2(0.12)=0.024$

$ln(128/25)=ln(125/25)+ Delta y$

$ln(128/25) ~~ ln(125/25)+ dy~~1.609+0.024=1.633$

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

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