Adriana Ayers

2022-06-26

Why does the method to find out log and cube roots work?

To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.

1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$

2)next, subtract 1, divide by 3 (for cube root, and any number n for nth root), add 1.

3) Then square the resultant number (say c) 13times (or 10 times if you had taken out root 10 times) i.e. ${c}^{{2}^{{2}^{....2}}}={c}^{{2}^{13}}$. This yields the answer.

I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.

For finding the log, the method is similar:-

1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.

To find cube roots of any number with a simple calculator, the following method was given to us by our teacher, which is accurate to atleast one-tenths.

1)Take the number $X$, whose cube root needs to be found out, and take its square root 13 times (or 10 times) i.e. $\sqrt{\sqrt{\sqrt{\sqrt{....X}}}}$

2)next, subtract 1, divide by 3 (for cube root, and any number n for nth root), add 1.

3) Then square the resultant number (say c) 13times (or 10 times if you had taken out root 10 times) i.e. ${c}^{{2}^{{2}^{....2}}}={c}^{{2}^{13}}$. This yields the answer.

I am not sure whether taking the square root and the squares is limited to 10/13 times, but what I know is this method does yield answers accurate to atleast one-tenths.

For finding the log, the method is similar:-

1)Take 13 times square root of the number, subtract 1, and multiply by $3558$. This yield s the answer.

Paxton James

Beginner2022-06-27Added 25 answers

Let's use these classical formulae :

${e}^{x}=\underset{n\to \mathrm{\infty}}{lim}{(1+\frac{x}{n})}^{n}$

$\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}x=\underset{n\to \mathrm{\infty}}{lim}n({x}^{1/n}-1)$

to get (replacing the limit by a large enough value of $n$:$N={2}^{13}$) :

$\begin{array}{rl}\sqrt[3]{x}={e}^{(\mathrm{ln}x/3)}& \approx {(1+\frac{\mathrm{ln}x/3}{N})}^{N}\\ & \approx {(1+\frac{N({x}^{1/N}-1)}{3\phantom{\rule{thinmathspace}{0ex}}N})}^{N}\\ & \approx {(1+\frac{({x}^{1/N}-1)}{3})}^{N}\end{array}$

Concerning the decimal logarithm we have :

${\mathrm{log}}_{10}\phantom{\rule{thinmathspace}{0ex}}x=\frac{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}x}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}\approx \frac{N}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}({x}^{1/N}-1)$

For $N={2}^{13}$ we may (as indicated by peterwhy) approximate the fraction with

$\frac{N}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}=\frac{{2}^{13}}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}\approx 0.4343\times 8192\approx 3558$

Hoping this clarified things,

${e}^{x}=\underset{n\to \mathrm{\infty}}{lim}{(1+\frac{x}{n})}^{n}$

$\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}x=\underset{n\to \mathrm{\infty}}{lim}n({x}^{1/n}-1)$

to get (replacing the limit by a large enough value of $n$:$N={2}^{13}$) :

$\begin{array}{rl}\sqrt[3]{x}={e}^{(\mathrm{ln}x/3)}& \approx {(1+\frac{\mathrm{ln}x/3}{N})}^{N}\\ & \approx {(1+\frac{N({x}^{1/N}-1)}{3\phantom{\rule{thinmathspace}{0ex}}N})}^{N}\\ & \approx {(1+\frac{({x}^{1/N}-1)}{3})}^{N}\end{array}$

Concerning the decimal logarithm we have :

${\mathrm{log}}_{10}\phantom{\rule{thinmathspace}{0ex}}x=\frac{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}x}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}\approx \frac{N}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}({x}^{1/N}-1)$

For $N={2}^{13}$ we may (as indicated by peterwhy) approximate the fraction with

$\frac{N}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}=\frac{{2}^{13}}{\mathrm{ln}\phantom{\rule{thinmathspace}{0ex}}10}\approx 0.4343\times 8192\approx 3558$

Hoping this clarified things,

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