Two main facts are needed here:
1. The logarithm [tex]\log x[/tex], regardless of the base of the logarithm, exists for [tex]x>0[/tex].
2. The square root [tex]\sqrt x[/tex] exists for [tex]x\ge0[/tex].
(in both cases we're assuming real-valued functions only)
By (2) we know that [tex]\sqrt{2x-1}[/tex] exists if [tex]2x-1\ge0[/tex], or [tex]x\ge\dfrac12[/tex].
By (1), we know that [tex]\log(\sqrt{2x-1}+3)[/tex] exists if [tex]\sqrt{2x-1}+3>0[/tex], or [tex]\sqrt{2x-1}>-3[/tex]. But as long as the square root exists, it will always be positive, so this condition will always be met.
Ultimately, then, we only require [tex]x\ge\dfrac12[/tex], so the function has domain [tex]\left[\dfrac12,\infty)[/tex].
To determine the range, we need to know that, in their respective domains, [tex]\sqrt x[/tex] and [tex]\log x[/tex] increase monotonically without bound. We also know that [tex]x=\dfrac12[/tex] at minimum, at which point the square root term vanishes, so the least value the function takes on is [tex]\log3[/tex]. Then its range would be [tex][\log3,\infty)[/tex].
