Can you hand calculate an unknown exponent above 0 and below 1 easily?

Question

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.

Answer 1

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 $10^k$.

Answer 2

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 $\approx 0.8$.

Answer 3

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.

Answer 4

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.)