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
`√(2x-1)` exists if
`2x-1\ge0`, or
`x\ge\frac12`.

By (1), we know that
`\log(√(2x-1)+3)` exists if
`√(2x-1)+3>0`, or
`√(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\frac12`, so the function has domain
`\left[\frac12,\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=\frac12` 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)`.