Question

What’s the domain and range of: log(√(2x-1) + 3 ) Please explain how you got it too!!

Answer 1

Two main facts are needed here:

1. The logarithm $\log x$, regardless of the base of the logarithm, exists for $x>0$.

2. The square root $\sqrt x$ exists for $x\ge0$.

(in both cases we're assuming real-valued functions only)

By (2) we know that $\sqrt{2x-1}$ exists if $2x-1\ge0$, or $x\ge\dfrac12$.

By (1), we know that $\log(\sqrt{2x-1}+3)$ exists if $\sqrt{2x-1}+3>0$, or $\sqrt{2x-1}>-3$. 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 $x\ge\dfrac12$, so the function has domain $\left[\dfrac12,\infty)$.

To determine the range, we need to know that, in their respective domains, $\sqrt x$ and $\log x$ increase monotonically without bound. We also know that $x=\dfrac12$ at minimum, at which point the square root term vanishes, so the least value the function takes on is $\log3$. Then its range would be $[\log3,\infty)$.