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## FAQs & How-to's

#### What is this calculator for?

This calculator takes the base e logarithm of a number.

#### Can I embed this on my website?

Sure. Embedding is allowed as long as you promise to follow our conditions. Here's the embed code:

 <iframe width="415" height="135" src="http://www.a-calculator.com/logarithm/embed.html" frameborder="0" allowtransparency="true"></iframe> 

#### How can I find the natural logarithm without a calculator?

In general, there isn't an exact way to do this. Manually approximating the log of a number is hard! To make it easier, we must remember three things. Firstly, that $$\label{series} \tag{1} \ln(x) = 2 \left( k + \frac{k^3}{3} + \frac{k^5}{5} + \ldots \right) \qquad \text{where } k = \frac{x-1}{x+1},$$ secondly, $$\ln(10) \approx 2.30 \qquad \text{(3 significant figures)},$$ and lastly we must remember our log laws!

Now, suppose we want to find an approximation of $$\ln(x_0)$$, then we would rewrite it as \begin{equation*} \ln(x_0) = \ln(x_0/10^n) + n \times \ln(10) \qquad \text{where } \frac{x_0}{10^n} \approx 1. \end{equation*} After rewriting it we can figure out $$\ln(x_0/10^n)$$ using equation $$\eqref{series}$$, and we know that $$\ln(10) \approx 2.30$$. The reason we want $$x_0/10^n$$ to be close to $$1$$ is so that we can use fewer terms from equation $$\eqref{series}$$ to get a good approximation (faster convergence).

As an example, if we want to find $$\ln(643)$$ we would do the following \begin{align*} \ln(643) &= \ln(0.643 \times 10^3) \\ &= \ln(0.643) + \ln(10^3) \\ &= \ln(0.643) + 3 \ln(10) \\ &\approx \ln(0.643) + 3 \times 2.3. \tag{2} \label{approx1} \end{align*}

To get $$\ln(0.643)$$ using $$\eqref{series}$$, we write $$k = (0.643 - 1)/(0.643 + 1) \approx -0.217$$ so that \begin{align*} \ln(0.643) & \approx 2 \left( -0.217 + \frac{(-0.217)^3}{3} + \frac{(-0.217)^5}{5} \right) \\ & \approx -0.441. \tag{3} \label{approx2} \end{align*} Notice that we've only had to write out the first three terms of equation $$\eqref{series}$$ to get an acceptable approximation in this case.

Substituting the result from equation $$\eqref{approx2}$$ back into $$\eqref{approx1}$$ we get \begin{align*} \ln(643) & \approx -0.441 + 3 \times 2.30 \\ & \approx 6.5, \end{align*} which is pretty close! In fact, using the calculator on this page we get $$\ln(643) \approx 6.46614$$.