# How to find antilog with simple calculator?

I know how to find log to base $10$ using simple calculator:

say if you want to find log of $12$ you can do as blow:

Step 1: $13$ times $\sqrt{\star} \implies 1.00030338$;
Step 2: subtract 1: $1.00030338 - 1 = 0.00030338$;
Step 3: Multiply by $3558 = 1.07942$.

Can I find $\operatorname{antilog}$ too?

By simple calculator i mean this:

• Define "simple calculator". Compared to mathematica, even a TI-83 is "simple". – Hayden Feb 7 '15 at 16:58
• Isn't $\operatorname{antilog}_{10} 12 = 10^{12}$. Am I missing something here? – rubik Feb 7 '15 at 17:06
• @rubik 12 is just an example. – Freddy Feb 7 '15 at 17:13
• @Hayden i have edited question. – Freddy Feb 7 '15 at 17:24
• Hey @Freddy, can you provide a link where I could check whether that log thing is real, not by just some examples but for all numbers? – Vicrobot Feb 25 '19 at 12:52

Using your calculator's square root key, you can approximate any antilog as closely as you like. Say you want $10^{1.234}$. You start by writing $$1.234 \approx 1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{128}$$ which you can find by any of several straightforward methods. (If this isn't clear, leave a comment and I will explain it further.)
Then you can calculate \begin{align}10^{1/8} &= \sqrt{\sqrt{\sqrt{10}}}\\ 10^{1/16} &= \sqrt{10^{1/8}}\\ 10^{1/32} &= \sqrt{10^{1/16}}\\ 10^{1/128} &= \sqrt{\sqrt{10^{1/32}}} \\ \end{align}
and so on. Then $$10^{1.234} \approx 10\cdot 10^{1/8}\cdot 10^{1/16}\cdot 10^{1/32} \cdot 10^{1/128}.$$
You can be a little more clever than this. $1.234$ is almost, but not quite, $1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{64}$, so you can get a much better approximationby writing $$1.234\approx 1 + \frac18 + \frac1{16} + \frac1{32} + \frac1{64} \color{red}{- \frac1{512}}$$
and then $$10^{1.234}\approx 10\cdot 10^{1/8}\cdot 10^{1/16}\cdot 10^{1/32}\cdot 10^{1/64}\color{red}{\div 10^{1/512}}.$$