Answer
a.) $e^{3y} = 5$
b.) $e^{-1} = (t+1)$
Work Step by Step
$Express$ $the$ $equation$ $in$ $exponential$ $form:$
a.) $\ln 5 = 3y$
b.) $\ln (t+1) = -1$
Natural Log form to exponential form: $\ln a = b \rightarrow e^b = a$ [Note: e means euler's number, NOT a variable]
a.) $\ln 5 = 3y$
Plug in 5 for a and 3y for b
$\ln 5 = 3y \rightarrow e^{3y} = 5$
b.) $\ln (t+1) = -1$
Plug in (t+1) for a and -1 for b
$\ln (t+1) = -1 \rightarrow e^{-1} = (t+1)$
