# How do you find h(4+t) given h(t)=2*3^(t+3)?

Mar 7, 2017

See the entire solution process below:

#### Explanation:

To solve this, substitute $\textcolor{red}{4 + t}$ for every occurrence of $\textcolor{b l u e}{t}$ in the function from the problem:

$h \left(\textcolor{b l u e}{t}\right) = 2 \cdot {3}^{\textcolor{b l u e}{t} + 3}$ becomes:

$h \left(\textcolor{red}{4 + t}\right) = 2 \cdot {3}^{\textcolor{b l u e}{\textcolor{red}{4 + t}} + 3}$

$h \left(\textcolor{red}{4 + t}\right) = 2 \cdot {3}^{4 + 3 + t}$

$h \left(\textcolor{red}{4 + t}\right) = 2 \cdot {3}^{7 + t}$