“Coincidence?” said the Mathematical Ninja. “I think not!” He looked at his whiteboard pen as if wondering how best to fashion a weapon out of it. He wrote:

$10^3 = 1,000$

$2^{10} = 1,024 = 1.024 \times 10^3$

So $10 \ln (2) = 3 \ln(10) + \ln(1.024)$

But $\ln(1.024) \simeq 0.024$

$10 \ln (2) \simeq 3 \ln(10) + 0.024$ - now divide by $\ln(10)$ to get everything in base 10:

$10 \log_{10}(2) \simeq 3 + \frac{0.024}{\ln(10)}$

“Isn’t $\ln(10)$ about $2.3$?” asked the student, brightly.

The Mathematical Ninja nodded. “So that last term is about $0.01$.”

“Which means $10 \log_{10}(2) \simeq 3.01$ and $\log_{10}(2) \simeq 0.301$! Are there any other good approximations you have handy?”

“Well… $\log_{10}(3) \simeq 0.477$, which is either $\frac{10}{21}$ (within 0.2%) or $\frac{43}{90}$ (within 0.16%).”

“And I suppose $\log_{10}(4)$ is just double $\log_{10}(2)$… because it’s $\log_{10}(2^2)$”, he clarified, catching a ‘just checking’ look. “So that’s 0.602, or $\frac{3}{5}$. Oo! And $\log_{10}(5)$ is $\log_{10}(10) - \log_{10}(2) \simeq 0.699$!”

“Very good,” said the Mathematical Ninja. “Meanwhile, $\log_{10}(6) = \log_{10}(2) + \log_{10}(3) \simeq \frac {3}{10} + \frac{43}{90} = \frac{7}{9}$.”

“That’s nice! So $6^9 \simeq 10^7$!”

“(To within 0.8%),” said the Mathematical Ninja. “The log approximation is good to 0.05%, though!”

“How about $\log_{10}(7)$?” asked the student. “I suppose I’d need to look for powers of $7$ which are close to nice multiples of powers of ten? Well, $7^2 \simeq \frac 12 \times 10^2$, so $2\log_{10}(7) \simeq 2 - \log_{10}(2) \simeq 1.7$… $\log_{10}(7) \simeq 0.85$?”

“Not bad,” said the Mathematical Ninja, “it’s 0.8451, give or take, which is $\frac{38}{45}$, although $\frac{76}{90}$ is easier to work with.”

“Ninetieths seem to be popping up a lot!”

“That,” said the Mathematical Ninja, “certainly is a coincidence.”

## 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.

## christianp

@icecolbeveridge I’ve got to take issue with your use of \simeq there. What’s wrong with \approx?

## ColinTheMathmo

RT @icecolbeveridge: [FCM] Why $log_{10} (2) \simeq 0.3$: http://t.co/rIsewYagGo

## ColinTheMathmo

.@icecolbeveridge It’s because 2^10 is about 10^3, yes?

## Nanette Amanda McLeod-Johnson

Nanette Amanda McLeod-Johnson liked this on Facebook.

## MathbloggingAll

Why $\log_{10} (2) \simeq 0.3$ http://t.co/E2BZAscSAE

## srcav

@icecolbeveridge the equations are broken! http://t.co/UxoUuJIkjs

## srcav

I enjoyed this from @icecolbeveridge http://t.co/UxoUuJIkjs