# The Mathematical Ninja and Logs Base 2

The student’s shoulder twitched slightly as he said “So I need to work out $\log_2(10)$…” and the crash of the cane against the table reminded him that the calculator was off-limits.

“I think you can estimate that yourself,” said the Mathematical Ninja.

“Uh… ok. There’s a change of base formula, isn’t there? So it’s the same as $\frac{\ln(10)}{\ln(2)}$. I don’t know $\ln(10)$ by heart, but it’s $\ln(5) + \ln(2)$, and I know $\ln(5)$ is 1.6.”

“1.609” murmurred the Ninja. “Same as miles to kilometres, the correct way to convert.”

“And I know $\ln(2)$ is 0.693. Ugh. But I can split it up and say my answers is $1 + \frac{1.609}{0.693}$, although that’s not exactly pleasant.”

“Suck it up, buttercup.”

“…? Just because you’ve got a cane, you think you can get away with anything.”

The Mathematical Ninja tilted their head. “Not *just* because I’ve got a cane.”

“Fine. I’m going to add 1% to both of those numbers to make the estimate easier. That turns it into $1 + \frac{1.625}{0.7}$, give or take.”

“I like your style.”

“The fraction bit is 16-and-a-quarter over 7, which is… 2… point… (uh, 22) 3 (ugh 15) 2. 2.32? So it works out to 3.32 in the end.”

“Not bad,” said the Mathematical Ninja.

### But there’s a better way

“It has to be a bit less than 3-and-a-third because $2^{10} \approx 10^3$. In fact, $3\ln(10) + \ln(1.024) = 10\ln(2)$.”

“And we can adjust that somehow?”

“Yes. We’re going to divide everything by $3\ln(2)$ to get $\ln_2{10} + \frac{\ln(1.024)}{3\ln(2)} = \frac{10}{3}$.”

“Ah! And $\ln(1.024)$ is pretty much 0.024, while $3\ln(2)$ is a smidge under 2.1, so we’re off by 0.01… can I say 0.011?”

“I reckon so.”

“So we get 3.322.”

“Correct to about one part in 50,000.”

“Good enough for me, sensei.”

* Edited 2020-07-27 to fix a typo before the Mathematical Ninja spot… OW! … and then again to fix it back. Oops.

## A selection of other posts

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