Square root of $\frac{2}{2^x}$; how do I find $x$?

Question

I have this:

$$\sqrt{\frac{2}{2^x}} = 9.313225746154785 \times 10^{−10}$$

(sqrt(2/(2^x)))

How should I find $x$? I know it's 61 for this case, but I'd like to know how to solve it for when I don't. Thanks!

Answer 1

Distribute the power through the fraction. $$\frac{\sqrt{2}}{2^{x/2}}=9.313225746154785\times10^{-10}$$ Rearrange fractions. $$\frac{\sqrt{2}}{9.313225746154785\times10^{-10}}=2^{x/2}$$ Take logarithms of both sides. $$\log_2\left(\frac{\sqrt{2}}{9.313225746154785\times10^{-10}}\right)=\log_2\left(2^{x/2}\right)$$ Remove $x$ from the logarithm on the right by the log rule for powers. $$x=2\log_2\left(\frac{2^{1/2}}{9.313225746154785\times10^{-10}}\right)$$ Rearrange with the log rules for fractions & multiplication. $$\therefore\quad x=21-2\log_2(9.313225746154785)+20\log_2(5)\approx61$$

This relies on the fact that $9.313\ldots\approx\left(\frac{5}{4}\right)^{10}$