## Intermediate Algebra (12th Edition)

We know that logarithms to base 10 are common logarithms, and $log_{10}x$ is equivalent to $log(x)$. Therefore, $log(6.3)=log_{10}6.3$. We know that for all positive numbers $a$ (where $a\ne1$), and all positive numbers $x$, $y=log_{a}x$ means the same as $x=a^{y}$. Therefore, $log_{10}6.3=x$ is equivalent to $10^{x}=6.3$. So, we know that $log_{10}6.3$ must be between 0 and 1, because $10^{0}=1$ and $10^{1}=10$.