# How do you evaluate ln 5?

Jun 17, 2016

You can use Newton's method to help a little.

#### Explanation:

$\ln 5$ is the zero of the function $f \left(x\right) = {e}^{x} - 5$

If you can calculate ${e}^{x}$ then you can calculate approximations for $\ln 5$ using Newton's method...

$f ' \left(x\right) = {e}^{x}$

Starting with an approximation ${a}_{0}$, iterate to get better approximations using the formula:

${a}_{i + 1} = {a}_{i} - \frac{f \left(x\right)}{f ' \left(x\right)} = {a}_{i} - \frac{{e}^{x} - 5}{e} ^ x = {a}_{i} - 1 + \frac{5}{e} ^ x$

As for ${e}^{x}$ you can use the Maclaurin series, which converges for any value of $x$:

e^x = sum_(k=0)^oo x^k/(k!)

I would be interested to know how well this method works using a truncated version of this series for ${e}^{x}$. How many terms for ${e}^{x}$ does it require to get a $5$ decimal place approximation for $\ln 5$?

I may dig a little deeper...

Using sum_(k=0)^9 x^k/(k!) as an approximation for ${e}^{x}$ and Newton's method, the approximation for $\ln 5$ I found was:

$\ln 5 \approx 1.6094454239$

Using sum_(k=0)^10 x^k/(k!) I found:

$\ln 5 \approx 1.6094389965$

Compare those with the proper value:

$\ln 5 \approx 1.6094379124$

If you do want to evaluate:

sum_(k=0)^10 x^k/(k!)

then it may be easier to use the form:

sum_(k=0)^10 x^k/(k!) = 1+x(1+x/2(1+x/3(1+x/4(1+x/5(1+

x/6(1+x/7(1+x/8(1+x/9(1+x/10)))))))))