Find the range of a logarithmic function whose domain is all real numbers between 2 and 10 (exclusive).

Question

$$h(x)=\log_{10}(x+1).$$

Shouldn't the range be all real numbers between $0.602$ and $1$ (inclusive)?

Answer 1

$$2<x<10 \implies 3<x+1<11 \implies \log_{10}3<\log_{10}(x+1)<\log_{10}11.$$ (The last step is due to the fact that $\log_{10}$ is an increasing function.) Therefore, the range of $h$ over $(2,10)$ is $(\log_{10}3, \log_{10}11)$, which is roughly $(0.47712, 1.04139)$.

Answer 2

For $1<x<2$ we have $$\log 3<\log (1+x)< \log(11)$$ Therefore the range is $$(\log 2, \log 11) = (.4771, 1.0413 )$$