How do you find $h (4 + t)$ $h(4+t)$ given $h (t) = 2 \cdot 3^{t + 3}$ $h(t)=2*3^(t+3)$ ?

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Answer 1 · 2017-03-07T23:21:03

See the entire solution process below:

To solve this, substitute $4 + t$ $color(red)(4 + t)$ for every occurrence of $t$ $color(blue)(t)$ in the function from the problem:

$h (t) = 2 \cdot 3^{t + 3}$ $h(color(blue)(t)) = 2 * 3^(color(blue)(t)+3)$ becomes:

$h (4 + t) = 2 \cdot 3^{4 + t + 3}$ $h(color(red)(4 + t)) = 2 * 3^(color(blue)(color(red)(4 + t))+3)$

$h (4 + t) = 2 \cdot 3^{4 + 3 + t}$ $h(color(red)(4 + t)) = 2 * 3^(4 + 3 + t)$

$h (4 + t) = 2 \cdot 3^{7 + t}$ $h(color(red)(4 + t)) = 2 * 3^(7 + t)$

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