"So that works out to be $10^{1.6}$," said the student, reaching for the calculator -- and, of course, recoiling as the Mathematical Ninja yelled "yeeha!" and lasso-ed it out of her hand.

"Forty," he said. "Too high by half a percent or so. 39.8."

The student paused. She would normally have checked, but her calculator was out of bounds, and apparently likely to remain so. "I'll have to take your word for it, sensei."

"You will," said the Mathematical Ninja. "Or we could work it through together, what do you think?"

"I think I don't have much option."

"Correct. You're aware that $\log_{10}(2) \approx 0.301$?"

The student's eyes moved from one side to the other and back again. "I... do now!"

"Well, $\log_{10}(40) = 2\log_{10}(2) + \log_{10}(10)$."

"Which is..." she thought for a moment. "1.602! But that's not exactly 1.6, is it? Is that where the half percent comes from?"

"Right," said the Mathematical Ninja. "It's 0.002 too high -- which, multiplied by $\ln(10) \approx 2.3$ -- is a bit less than half a percent."

"And half a percent of 40 is 2, so you took that off. Smart. Let me think for a moment."

The Mathematical Ninja smiled, and cracked the lasso absent-mindedly around the room, picking up a slide rule, a book about topology and a small stuffed sheep.

"Does that mean," asked the student, carefully, "you can estimate ten to the power of any... one decimal place number pretty accurately? If you're starting with $10^{0.3} \approx 2$, give or take a quarter-percent, you could get to, I don't know, $10^{3.1}$ by doing $10^1 \times 10^{2.1}$, which is about $10 \times 2^7$, which is 1280 -- although you're off by about seven-quarters of a percent, let's say 20? So I'd go for 1260."

"1258.9," said the Mathematical Ninja. "But good effort."

* Edited 2015-09-28 to correct an operation. Thanks, Hannu!