# Secrets of the mathematical ninja: Estimating natural logs

Natural logs are just about the easiest part of the A-level syllabus to look like a god in -- because they're wrongly seen as difficult. In fact, once you know a handful of tricks, you can rattle off things such as 'log of 12 is about 2.5, of course...' (it's 2.485) with confidence and aplomb.

Here are a few key values to know by heart:

• $\ln(1) = 0$ (exact)
• $\ln(2) = 0.7$ (-1%)
• $\ln(3) = 1.1$ (-0.1%)
• $\ln(5) = 1.6$ (+0.6%)

And a few rules you need to know whether you want to be a ninja or not:

• $\ln(ab) = \ln(a) + \ln(b)$
• $\ln(a^b) = b \ln(a)$
• $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$

So, if you want to know $\ln(100)$, you can say 'that's $2 \ln (10)$', and then '$\ln(10)$ is $\ln(2) + \ln(5)$, which is $0.7 + 1.6$, about $2.3$ -- so $\ln(100)$ is about 4.6'. (It's 4.605).

There's also a really helpful approximation for natural logs of numbers near 1:

$$\ln(1 + x) \approx x$$

That goes deeper than it looks. If you're looking for the log of a number that's slightly off something you know -- say you want to know $\ln(17)$ for some reason, and you spot straightaway that $\ln(16)$ is $4 \ln(2)$, or 2.8 (less 1%, which is 2.77). You can say 'I'd need to add about 6% (of 16) on to 16 to get 17 -- so I need to add 0.06 on to $\ln(16)$ to get $\ln(17)$, which will make it 2.83. (It's 2.833).

But wait -- there's more. If you're smart with your dividing, you can solve the silly powers questions they give you in C2, such as $5^x = 17$.

When you rearrange that, you get $x = \log_5 (17)$, which is the same as $\frac{\ln(17)}{\ln(5)}$. You know both of those: it's $\frac{2.83}{1.6}$. Rounding that off to $\frac{2.8}{1.6}$ or $\frac{28}{16}$, you get an answer of $\frac{7}{4}$, which is 1.75. (The actual answer is 1.76 -- which isn't at all shabby without a calculator).

(There's a serious point to this: if you can work this natural logs stuff out in your head, you can look at your calculator and say 'that looks ok' or 'that looks dubious'. Looking ok isn't a guarantee that you have the right answer, but looking dubious is a good sign that you need to check your work more carefully and figure out whether your mistake is in the estimate or the sum.)

## Colin

Colin is a Weymouth maths tutor, author of several Maths For Dummies books and A-level maths guides. He started Flying Colours Maths in 2008. He lives with an espresso pot and nothing to prove.

### 6 comments on “Secrets of the mathematical ninja: Estimating natural logs”

• ##### pozorvlak

I recently came across a rather wonderful approximation for natural logarithms, from https://www.johndcook.com/blog/2012/11/24/approximation-relating-lg-ln-and-log/:

ln(x) =~ log_2(x) – log_10(x)

This is especially useful if you know a few powers of 2, as many computery people do. So, for instance, ln(1000) =~ log_2(1000) – log_10(1000) =~ 10 – 3 = 7 (the actual value is 6.91 to 3sf).

• ##### Colin

Oh, that’s VERY nice!

• ##### Heba

Hi Colin ,

I want a good approximation for sum natural logarithm in stead of factorial

• ##### Colin

I’m not sure exactly what you’re asking – what are you trying to sum? (It sounds like Sterling’s Approximation might be your friend, but it’s hard to tell from your question.)

• ##### Maxim Winter

My favorite method for accurately approximating the ln(n) is to approximate the definite integral from 1 to n of x^-1 using the Trapeziodal Rule. This works because the antiderivative of x^-1 is ln(abs(x)). You can be extremely accurate by splitting up the integral into multiple intervals, and this doesn’t take too long because the Trapeziodal Rule is easy and quick to use: the integral from a to b of f(x) is approximately (b-a)*(f(a)+f(b))/2. Just substitute 1 for a, n for b, and x^-1 for f(x). This method is not quite as quick as some of the other methods mentioned above, but this method can be very accurate without having to use any guess and check, and you don’t have to remember any other natural logs (besides ln(1)=0). Best of all, in theory, if one split the integral infinitely and used the Trapeziodal Rule for the infinitely small individual integrals (and then add them together), you would obtain the exact natural logarithm of n.

• ##### Colin

Oo, nice idea! So $\ln(4)$ is about $\frac{1}{2} \left( 1 + \frac{1}{4} + 2\times\left(\frac{1}{2} + \frac{1}{3}\right)$.

That’s half of (five-quarters plus five-thirds), or $\frac{35}{24}$. I make that 1.46, which is not *bad* for a first attempt. I like it!

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