Written by Colin+ in logarithms, ninja maths.

You might think 'when would I ever write to an abstract concept like logarithms', but you'd be surprised. I had occasion to write this letter a few weeks ago:

Dear Base 10 Logarithms,

I'm very sorry that I bullied you. I recall saying things like 'grown-ups always use natural logs', and that your button ought to be expunged from calculators. I was wrong to say those things, which I believed to be true; I have now changed my mind and wanted to try to patch things up between us.

Since saying those terrible, ugly words - frankly, I'm ashamed of them - I've discovered your tremendous power* in the world of Ninja Maths. It's to your credit that you never got defensive about the torrents of abuse coming from me - and others - to say 'but look at standard form!' You kept your dignity and poise throughout. I salute you.

With kindest, humblest regards,

The Ninja Mathematician

* No pun intended!

Now, there's a hint in there about why I felt compelled to try to make my peace with $\log_{10}$ - and it's all to do with standard form. You remember playing with that at GCSE, right? Where you've got a number between one and ten, and then ten to the power of something? That makes it easier to work with very big and very small numbers instead of writing out huge strings of zeros.

Something you never do with standard form - and you totally should - is take logs of it. For instance, the log of $5 \times 10^4$ is - without looking at a table or a calculator - about 4.70. (With a calculator, it's 4.698: I was 0.02% off.)

How did I get that? Well, I'm starting to learn my logs base 10. $\log_{10}(5)$ is very close to 0.7, and $\log_{10}( 5 \times 10^4 ) = \log_{10}(5) + \log_{10}(10^4) = 4 + \log_{10}(5)$, or 4.7.

If I wanted to square root that - and who wouldn't? - I'd just have to halve the log to get 2.35, and work out $10^{2.35}$. Well... that's harder. I know it's between 100 and 1000, because it's between $10^2$ and $10^3$, but after that I can ignore the 2 completely. What's $10^{0.35}$? I'm going for 2.23, but only because I know my square roots. (Doing it blindly, I'd say $10^{0.3}$ is 2, another one I've learnt; 'a bit more than 200' would be a more honest guess.)

How about the tenth root? That, I have no idea about just looking at the numbers. However, I can say $4.7 \div 10$ is 0.47 - and I know that $10^{0.477}$ is 3. I should correct that downwards slightly - maybe 2.95? There we go.

As long as it's a positive number, the number before the decimal point is the power of 10, and working out 10 to the power of the other bit gives you the number to multiply by.

Here's a quick table to learn, if you're into that kind of thing:

$x$ | $log_{10}(x)$ | comment |
---|---|---|

1 | 0.000 | by definition! |

2 | 0.301 | very close to 0.3 |

3 | 0.477 | 0.5 (-2.3%) |

4 | 0.602 | double $\log_{10}(2)$ |

5 | 0.699 | very close to 0.7 |

6 | 0.778 | very close to 7/9 |

7 | 0.845 | 5/6 (+1.4%) |

8 | 0.903 | Three times $\log_{10}(2)$ |

9 | 0.954 | Double $\log_{10}(3)$ |

10 | 1.000 | By definition! |

What other creative ways of using logs base ten can you come up with?

Want to know more about handling logarithms for C2?

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