Can you hand calculate an unknown exponent? I was recently calculating something and entered $\log 6.7$ in my calculator only to quickly feel frustrated that I did not even know how to begin to calculate by hand the number. Is there a method to hand calculate sort of like long division? Usually I just use my mind to calculate such a problem(integers) but because the answer is a bunch of random decimals I need some method to hand calculate it.

I understand divide $10$ to go up and multiply $10$ to go down... $$\begin{array}{c} 10^{-2} &=& \frac{1}{100} \\ 10^{-1}& =& \frac{1}{10} \\ 10^{0}& = &1 \\ 10^{1} &= &10\\ 10^{2} &=& 100\\ \end{array}$$

But I don't quite seem to understand the transition from exponent $1$ to exponent $0$ for example $10^{.5} = 3.16227766017...$ I don't really understand why it equals $3.16$ and not $5$ and yet $100^{.5} = 10.$

So far I got $10^?=6.7$

I logically know it is in between $1$ and $0$ but have no idea how to calculate it on paper.

  • $\begingroup$ Taking something to the $n$th power is not the same as multiplying by $n$, so there is no reason to expect $10^{1/2}$ to be $10/2$. In fact, $(10^{1/2})^2=10$ so $10^{1/2}$ must be the square root of $10$, which is approximately $3.16$. As for the equation $10^?=6.7$, that involves logarithms. Ultimately, doing roots, fractional powers and logarithms are not easy by hand, and practically impossible without any grasp of analysis and numerical methods. $\endgroup$ – anon Jul 29 '13 at 8:04
  • $\begingroup$ The best reference for practical calculation I know is Volume 1 of the Feynman Lectures on Physics (I think Chapter 22). $\endgroup$ – Mark Bennet Jul 29 '13 at 8:08
  • $\begingroup$ @anon (10^1/2)^2, this has really helped me understand where to begin hand calculating the answer. I didn't realize an exponent of 1/2 is equal to the square root and 1/3 is the cubed root... now I just need a quick way of estimating large roots and it should only take a few minutes to brute force a few decimal places... although not exactly what I was looking for... $\endgroup$ – CodeCamper Jul 29 '13 at 8:18

One way to find an approximation $\frac ab$ for $\log 6.7$ wuld be to find an integer $b$ such that $6.7$ to the $b$th power is approximately a power of $10$ (say, $10^a$). Then from $10^a\approx 6.7^b$ we get $a=\log(10^a)\approx\log(6.7^b)=b\log 6.7$ and hence $\log 6.7\approx \frac ab$. But of course this is laborious because it is a piece of luck to hit approximately a power of $10$. For example $6.7^2\approx 45$ is far from a power of ten - almost right between $10^1$ and $10^2$, which gives us only the rough estimate $\frac12<\log 6.7<1$. And $6.7^3\approx 300$ is not much better, giving only $\frac23<\log6.7<1$. However, from $3^2\approx 10$ we see that $6.7^6\approx 300^2\approx 100000$ and hence $\log 6.7\approx \frac56$.

Some easy results may help making good approximations. For example $2^{10}=1024\approx 10^3$, so $\log 2\approx 0.3$. Likewise (and essentially we just used this fact above), $\log 3\approx0.5$. Thus $\log 6=\log 2+\log 3\approx 0.8$. Now $6.7$ is just $10\%$ larger, so more refintement might not help as we have already plugged in two approximations with small but not negligible error.

Another way, again using only $\log 2\approx 0.3$ and $\log 3\approx 0.5$, is to note that $3\cdot 6.7\approx 20$, so $\log 3+\log 6.7\approx1+\log 2$, which again gives us $\approx 0.8$.


Other answers have discussed about approximation of $\log$, so I will address something else you may have been confused about: raising a number to a non-integer power. (This may not seem like an answer, but it seems too long to be a comment.)

To clear the confusion, you mentioned $10^{0.5} \ne 5$ but $100^{0.5} = 10$. I understand that you might imagine $$ 2 \times 10^{0.5} = 100^{0.5} = 10. $$ But the left equality here is not true. It is true, however, that $$ (10^{0.5})^2 = (10^2)^{0.5} = 10 = 100^{0.5}. $$ So, essentially, $2 \times 10^{0.5}$ is not the same as $10^{0.5} \times 10^{0.5}$. I hope that this, in a way, explains why $10^{0.5}$ is not $5$.

Raising a number to the $a$-th power with $a$ between $0$ and $1$ is not as simple as taking the average in the usual sense. As an example above showed, $10^{0.5}$ has to be a number such when you square it, you get $10$. (Because $(10^{0.5})^2 = 10^{0.5} \times 10^{0.5} = 10^{0.5 + 0.5} = 10^1 = 10$.) In fact, $10^{0.5}$ is the square root of $10$. I believe you know how to calculate an approximation of $\sqrt{10}$ by hand.

You can elaborate this idea further. Suppose you are told to compute an approximation of $10^{0.83}$. You can use a sequence of square root operations and multiplications to accomplish this. You can approximate $0.1$ using the binary expansion $0.83_{10} = 0.110101\ldots_2$, i.e., $$ 0.83 = 2^{-1} + 2^{-2} + 2^{-4} + 2^{-6} + \ldots $$ Then, $10^{0.83}$ can be approximated by $$ 10^{0.83} = 10^{2^{-1} + 2^{-2} + 2^{-4} + 2^{-6} + \ldots} = 10^{2^{-1}} \times 10^{2^{-2}} \times 10^{2^{-4}} \times 10^{2^{-6}} \times \ldots $$ $10^{2^{-n}}$ can be obtained by computing the square root of $10$ $n$ times. The procedure to compute an approximation of $10^{0.83}$ goes as follows.

First, compute $x_1 = 10^{2^{-1}} = \sqrt{10} \approx 3.16$. Next, compute $x_2 = \sqrt{x_1} \approx 1.78$. Keep repeating this way to get $x_3 = \sqrt{x_2} \approx 1.33$, $x_4 = \sqrt{x_3} \approx 1.15$, $x_5 = \sqrt{x_4} \approx 1.07$, $x_6 = \sqrt{x_5} \approx 1.03$. Let's pause here and compute $x_1 \times x_2 \times x_4 \times x_6 \approx 3.16 \times 1.78 \times 1.15 \times 1.03 \approx 6.66$. This is our approximation of $10^{0.83}$ using only the first six digits of the binary expansion of $0.83$, and only two decimals for each factor.

Wolfram Alpha gives $10^{0.83} \approx 6.76$, so our approximation is off by $0.10$. The error can be reduced by adding more digits in each step and adding more terms from the binary expansion of $0.83$.

(There are many other methods to compute a non-integer power of a number, and none of them are very easy.)


well, logarithmic tables had costed (human) calculators a lot of labour, having to work out square roots with a lot of digits, and then multiplications and interpolations for finding the needed values with the desidered approximation. This is why there vere logarithmic tables, indeed :-)

If you just want to have a rough approximation, you should rote learn the logarithms of some small number. In your case, $6.7 \approx \frac{20}{3}$, so $\log 6.7 \approx \log 20 - \log 3 \approx 1.301 - 0.477 = 0.824$.

[EDIT] if you want a fairer approximation, you should (a) start with more digits, so that you get $\log 6.6667 \approx 1.30103 - 0.47712 = 0.82391$, and (b) apply linear interpolation. $\log 6 = \log 3 + \log 2 \approx 0.77815$ and $\log 7 \approx 0.84510$ (btw, I just memorized $\log 2$, $\log 3$ and $\log 7$). So the difference between $\log 6$ and $\log 7$ is roughly 0.067; the difference between 6.6666 and 6.7 is 0.0333, and therefore you should add 0.002+, obtaining 0.826. The actual value is 0.82607+, so the approximation is rather good.


For more precision, you can approximate the log with rational numbers. For example, if you know your powers of $6.7$, then you can observe that:

  • $\log6.7=\dfrac{1}{5}\log6.7^5=\dfrac{1}{5}\log13501.25107>\dfrac{1}{5}\log10000=\dfrac{4}{5}=0.8$
  • $\log6.7=\dfrac{1}{6}\log6.7^6=\dfrac{1}{6}\log90458.382169<\dfrac{1}{6}\log100000=\dfrac{5}{6}=0.833...$

So we know that $0.8 < \log 6.7 < 0.833...$. Unfortunately, these types of computations can quickly get super tedious to do by hand; basically, to get an accurate approximation, you want to find some positive integers $j,k$ such that $6.7^j$ is just under or just over $10^k$.


$\log 6.7$ means "what power do I need to raise 10 to in order to get 6.7?".

Put in another way: $10^{\log 6.7}=6.7$.

Yet another way: we have $10^x=6.7$, and the answer is $x=\log 6.7$.

Normally we can look up $\log 6.7$ in a log table or use a calculator to see what the decimal number version of $\log 6.7$ is. But what if we couldn't do that?

Okay, what if we do this:

$10^0=1$ and $10^1=10$.

$10^x=6.7$, so $x$ must be between 0 and 1.

Let's take the average: $\frac{0+1}{2}=\frac{1}{2}$. Taking the average is easy and it ensures that we end up somewhere "between" the values we're taking the average of (0 and 1 in this case).

$10^\frac{1}{2}\approx 3.16$. Okay so less than 6.7.

Let's take the average again: $\frac{1/2+1}{2}=\frac{3}{4}$

$10^\frac{3}{4}\approx 5.62$. Still less than 6.7

Let's take the average again: $\frac{3/4+1}{2}=\frac{7}{8}$

$10^\frac{7}{8}\approx 7.5$. So now we're over 6.7. Let's just take the average of $3/4$ and $7/8$ then.

Let's take the average again: $\frac{3/4+7/8}{2}=\frac{13}{16}$.

$10^\frac{13}{16}\approx 6.49$. Less than 6.7, but we're getting closer.

If we keep doing this eventually we'll get a number such that $10^{\text{number}}=6.7$. Implementing this method in Python, it took 26 iterations to get $10^{0.82607480\;13854027}=6.69999997970653$.

Wolfram alpha says $\log_{10} 6.7 = 0.82607480\;27...$.

If we wanted to do this by hand, we'd have to assume we had a method for approximating numbers like $10^\frac{13}{16}$ of course.

So just on the face of it, it seems like a lot of work to calculate logs between 0 and 1 by hand. There are better ways to approximate logs then what I've chosen in this post, but they all rely on doing a lot of calculation by hand (if you don't have a computer).

But doing a heck load of manual computations was how the first log tables were made. Napier spent 20 years making his. Log tables give you a faster and more reliable way to do multiplication - but only because someone went through the massive effort of putting together a log table in the first place.


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