How to get to this result? $\sqrt{\frac{2\times6.73\times10^{-19}}{9.109\times10^{-31}}}=\sqrt{1.50\times 10^{12}}$

Question

I'm sorry but I can't understand what's happening between $$\sqrt{\frac{2\times6.73\times10^{-19}}{9.109\times10^{-31}}}=\sqrt{1.50\times 10^{12}}$$ This is what the solutions from my manual have written on them, but I don't get how that operation was done. Thanks in advance

Answer 1

We have $6.73 = 673\times 10^{-2}$ and $9.107=9107\times 10^{-3}$ and recall that $10^a\times 10^b=10^{a+b}$ for $a,b\in\mathbb{R}$.

So, we get the following

$$\sqrt{\frac{2\times 673\times 10^{-2}\times 10^{-19}}{9107\times 10^{-3}\times 10^{-31}}}= \sqrt{\frac{2\times 673\times 10^{-21}}{9107\times 10^{-34}}}= \sqrt{\frac{2\times 673\times 10^{34} \times 10^{-21}}{9107}}=\sqrt{0.1477\times 10^{13}}$$

which leads to approximately $\sqrt{1.5\times 10^{12}}$ since $\frac{2\times 673}{9107}=0.1477\approx 0.15$.

Answer 2

For the exponents, one of the classic rules is $\frac {a^b}{a^c}=a^{b-c}$. This is applied here with $a=10, b=-19,c=-31$. The calculation with the rest results in about $1.47765946$, not $1.50$, so that part is wrong.

Answer 3

They are not equal!

$$\sqrt{\frac{2\times6.73\times10^{-19}}{9.109\times10^{-31}}}=1215590.16847 \tag1$$

$$\sqrt{1.50\times10^{12}}=1224744.87139\tag2$$

This is a difference of about $ABS(1215590.16847-1224744.87139)=9154.70292$

You could simplify the first root. Since: $$2\times\frac{6.73}{9.109}=1.47765945768 \tag3$$

and $$\frac{10^{-19}}{10^{-31}}=10^{\left(-19+31\right)}=10^{12}$$

You could write (1) as:

$$\sqrt{1.47765945768\times10^{12}}\tag 4$$

If you decided to round (3) so that you use $1.5$ instead of $1.47765945768$

You would get (4)=(2).