Answer
Makes sense.
Work Step by Step
By definition (see p.456),
$\log_{8}16$ is a number $x_{1}$ such that $8^{x_{1}}=16.$
Since the function $f(x)=8^{x}$ has base $8 > 1$, it is always increasing.
So, for x taken from x=1 to x=2, all the function values will be between
$f(1)=8$ and $f(2)=64.$
Any function value between 8 and 64 will be the function value of an x between 1 and 2.
$x_{1}$ is a number such that its function value is $f(x_{1})=16$, so
since $8 < 16 < 64$,
$x_{1}$=$\log_{8}16$ is somewhere between x=1 and x=2 .
