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:
Answer
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 $$10^{1.234}\approx 10\cdot 10^{1/8}\cdot 10^{1/16}\cdot 10^{1/32}\cdot 10^{1/64}\color{red}{\div 10^{1/512}}.$$
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