*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 colin@flyingcoloursmaths.co.uk 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

Hi, LEHMER!

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 $\log(2) \approx 0.30103$, 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

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.