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How does $h(t) = 8^t$ change over the interval from $t = 6$ to $t = 7$ ? $\newline$ Choices: $\newline$ (A) $h(t)$ increases by $800\%$ $\newline$ (B) $h(t)$ decreases by $8$ $\newline$ (C) $h(t)$ increases by $8\%$ $\newline$ (D) $h(t)$ increases by a factor of $8$

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Find derivatives of logarithmic functions

Full solution

Q. How does

h(t) = 8^t

change over the interval from

t = 6

to

t = 7

?

\newline

Choices:

\newline

(A)

h(t)

increases by

800\%

\newline

(B)

h(t)

decreases by

8

\newline

(C)

h(t)

increases by

8\%

\newline

(D)

h(t)

increases by a factor of

8

Calculate $h(6)$ : Calculate the value of $h(t)$ at $t = 6$ . $\newline$ $h(6) = 8^6$
Calculate $h(7)$ : Calculate the value of $h(t)$ at $t = 7$ . $\newline$ $h(7) = 8^7$
Find difference for change: Find the difference between $h(7)$ and $h(6)$ to determine the change. $\newline$ Change = $h(7) - h(6) = 8^7 - 8^6$
Factor out $8^6$ : Factor out $8^6$ to simplify the expression. $\newline$ Change = $8^6 \times (8 - 1) = 8^6 \times 7$
Determine factor of increase: Since $8^6$ is a common factor, the change is actually by a factor of $8$ . So, $h(t)$ increases by a factor of $8$ from $t = 6$ to $t = 7$ .

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