By definition, if
\(b^y = a\)
then
\(log_{b}a = y\)
In this case, if
\(4^{y} = 8\)
then
\(log_{4} 8 = y\)
We have:
\(4^{y} = 8\)
\(2^{2y} = 2^{3}\)
\(\diagup\!\!\!2^{2y} = \diagup\!\!\!2^{3}\)
\(2y = 3\)
\(y = \frac{3}{2}\) or \(y = 1.5\)
logaritmic graph (Follows the attachment):
