If I have a very large exponent (say $7^{2009}$), how can I quickly work out the first two digits of the number?

Thanks in advance.

  • $\begingroup$ Use logarithms. $\endgroup$ Mar 9, 2017 at 11:21
  • $\begingroup$ How would I go about doing that? $\endgroup$ Mar 9, 2017 at 11:21
  • $\begingroup$ Do you know what logarithms are? $\endgroup$ Mar 9, 2017 at 11:26
  • $\begingroup$ Yes, but I don't know how to use them for that. $\endgroup$ Mar 9, 2017 at 11:29
  • 2
    $\begingroup$ I take that to mean that you can't do that calculation. So, I suggest the first thing for you to do is to learn how to do that calculation. Then you can look up the terms "characteristic" and "mantissa", and you'll be well on your way to teaching yourself how to solve the problem. $\endgroup$ Mar 9, 2017 at 11:40

3 Answers 3


Let $A=7^{2009}$

Taking logarithm both the sides :

$\log_{10}A = \log_{10}7^{2009}$

$\log_{10}A = 2009\log_{10}7$

Put the value of $\log_{10}7$ now

$\log_{10}A \approx 1697.801962$

$A = 10^{1697.801962}$

$A = 10^{1697}×10^{0.801962}$

$A = 6.3381×10^{1697}$

Now you have got first two digits,

$6$ and $3$


We can use logarithms for this; first note that


Let's now set $n+b=2009\log_{10}(7)$, where $n\in\mathbb{Z}$ and $b\in(0,1)$. Now note that $$7^{2009}=10^{\log_{10}(7^{2009})}=10^{2009\log_{10}(7)}=10^{x+b}=10^n10^b$$ since $n$ is an integer and $b$ is between $0$ and $1$, we know $10^b$ is between $1$ and $10$. Thus, $10^b\cdot 10^n$ is simply the so-called "scientific notation" of $7^{2009}$. This includes an easy way to read off the first couple of digits; in this case, we have

$$b=0.801962388641...\text{ and } n=1697$$ and so $$10^b=6.3381481...$$ and as such $$7^{2009}=6.3381481...\cdot 10^{1697}$$ thus, the first couple of digits are $6338...$


If you're not familiar with the logarithm laws there's still some (slight) hope, but as you will see - you'd better learn the logarithm laws.

What we can do is to make the calculation approximately with enough precision to get two correct digits in the first place. To see what where we're going we have:

$$7^2 = 7\times 7 = 49$$ $$7^4 = 49\times 49 = 2401$$ $$7^8 = 2401\times 2401 = 5764801 \approx 576480\times 10^1 $$ $$7^16 = 7^8\times 7^8 \approx 576480^2 \times 10^2 = 332329190400\times 10^2 \approx 332329 \times 10^8 $$

Finally we may end up with (depending on details):

$$7^{2009} \approx 633413\times 10^{1692}$$

Here we note that this is not actually very near the number others got:

$$7^{2009} \approx 633815\times 10^{1692}$$

but it's close enough. The hard part with this approach, besides all the calculations is that you have to keep track on error estimates for each $\approx$ - and hopefully find that the result in the end is accurate enough to conclude that the first two digits are $6$ and $3$. Depending how well you do the estimates you can get by by using six digits in the calculations, but more digits makes the estimates easier.

What you need to do is to get an estimate of the error during multiplication. If $a(1+\epsilon_a)$ is the true value of the quantity expressed by $a$, $\epsilon_a$ being the relative error. You have:

$$a(1+\epsilon_a)\times b(1+\epsilon_b) = a\times b (1+\epsilon_a + \epsilon_b + \epsilon_a\epsilon_b)$$

We can for example knowing that $|\epsilon|<1$ estimate the relative error as $|\epsilon| < |\epsilon_a|+|\epsilon_b| + \min(|\epsilon_a|,|\epsilon_b|)$. Also we have in the rounding in each step an introduced error of $5\times 10^{-n}$ where $n$ is the number of digits. Note that by this way of estimating the error you will likely need more digits to get to a good enough error estimate to have a conclusion.


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