[tex]\begin{array}{llll} \textit{Logarithm of exponentials} \\\\ \log_a\left( x^b \right)\implies b\cdot \log_a(x) \end{array}~\hfill \begin{array}{llll} \textit{Logarithm Cancellation Rules} \\\\ \stackrel{\stackrel{\textit{we'll be using this one}}{\downarrow }~\hfill }{log_a a^x = x\qquad \qquad a^{log_a x}=x} \end{array}[/tex]
[tex]\textit{Logarithm Change of Base Rule} \\\\ \log_a b\implies \cfrac{\log_c b}{\log_c a}\qquad \qquad c= \begin{array}{llll} \textit{common base for }\\ \textit{numerator and}\\ denominator \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}[/tex]
[tex]2^{0.5x}=10\implies \log_2\left( 2^{0.5x} \right)=\log_2(10)\implies 0.5x=\log_2(10) \\\\\\ \cfrac{x}{2}=\log_2(10)\implies x = 2\log_2(10)\implies x = \log_2(10^2)\implies x = \log_2(100) \\\\\\ \stackrel{\textit{using the change of base rule}}{x = \cfrac{\log(100)}{\log(2)}\implies x = \cfrac{2}{\log(2)}}\implies x\approx 6.644[/tex]
as far as the domain, or namely what values "x" can take on safely, I don't see any constraints, so it must be (-∞ , +∞).
