Answer
$y\approx3.6439$
Work Step by Step
We are asked to solve:
$2^{y+1}=25$
Recall the definition of a logarithm (pg. 451):
$\log_{b}{x}=y$ iff $b^y=x$
Applying this definition to our equation (with $b=2, y=y+1, x=25$), we get:
$\log_{2}{25}=y+1$
Next, recall the change of base formula (pg. 464):
$\log_{b}{m}=\frac{\log_{c}{m}}{\log_{c}{b}}$
Applying this property to our last equation, we get:
$\log_{2}{25}=y+1$
$\dfrac{\log{25}}{\log{2}}=y+1$
Using a calculator, we get:
$4.6439=y+1$
$y=4.6439-1$
$y\approx3.6439$
