Ask Uncle Colin: A Huge Power Of Two

Ask Uncle Colin is a chance to ask your burning, possibly embarrassing, maths questions -- and to show off your skills at coming up with clever acronyms. Send your questions to and Uncle Colin will do what he can.

Dear Uncle Colin,

I've been asked to find $2^{64}$ without a calculator, to four significant figures. How would you go about this?

-- Large Exponent, Horrific Multiplication, Extremely Repetitive


To get a rough answer, I'd usually start with the rule of thumb that $2^{10} \approx 10^3$. I'd conclude that $2^{60} \approx 10^{18}$, so $2^{64} \approx 1.6 \times 10^{19}$, plus an adjustment... but that's not going to get close enough for this problem.

Option 2 is the binomial expansion, based on the above: $2^{64} = 16 \times (1024)^{6} = 16 \times 10^{18} \times (1.024)^6$. Now, $(1+x)^6 \approx 1 + 6x + 15x^2$, for small $x$, which works out to be $1 + 0.144 + .00864 = 1.15264$. I need to multiply this by 16, which I'd do by quadrupling (4.61056) and quadrupling again (18.44224), giving an approximation of $1.844 \times 10^{19}$. That's close (the real answer is 1.845)-- and we can close the gap by taking an extra term in the binomial expansion.

The juicy bit

A third option, my favourite, is to work in base-ten logs. If we know that $\log(2) \approx 0.30103$1, we can easily say that $\log\left(2^{64}\right) = 64 \log(2) \approx 19.26592$. This tells us that $2^{64}$ is a bit more than $10^{19}$.

Next, we need to deal with the decimal part. $10^{0.266}$ is trickier, to say the least! It's between 1 ($10^0$) and 2 ($10^{0.30103}$), for sure. $\log(1.8) = 2 \log(3) + \log(2) ~ 0.954 + 0.301 - 1= 0.255$, so the log correction between our number and $\log(1.8)$ is 0.011.

Since $10^{0.011} = e^{0.011 \ln(10)} \approx e^{0.0253} \approx 1.025$, we'd add on 2.5% to my estimate of 1.8 to get 1.845.
As noted before, $2^{64} \approx 1.845 × 10^{19}$, correct to 4sf.

I'm not sure Option 3 is really more accurate than option 2, in fairness. Without a calculator, I'd back option 2 to give a better answer most times; here, it got unlucky with the rounding.

Hope that helps!

-- Uncle Colin

* Edited 2016-10-12 to fix several LaTeX errors.


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.

  1. to about one part in 70 million, incidentally []


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