Answer
$\log_{49}{7}=\frac{1}{2}$
Work Step by Step
Let
$x=\log_{49}{7}.$
RECALL:
$\log_{a}{y} = b \longleftrightarrow a^b = y.$
Use the definition above to convert the equation to an exponential equation and have
$49^x=7.$
Write each side in base 7 to have
$(7^2)^x=7
\\7^{2x}=7.$
The bases are the same, so the exponents must be equal. Thus,
$2x=1
\\x = \frac{1}{2}.$
Therefore $\log_{49}{7}=\frac{1}{2}$.
