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How does h(x)=8xh(x) = 8^x change over the interval from x=1x = 1 to x=2x = 2?\newlineChoices:\newline(A) h(x)h(x) increases by 800%800\%\newline(B) h(x)h(x) increases by a factor of 88\newline(C) h(x)h(x) decreases by 88\newline(D) h(x)h(x) increases by x=1x = 100

Full solution

Q. How does h(x)=8xh(x) = 8^x change over the interval from x=1x = 1 to x=2x = 2?\newlineChoices:\newline(A) h(x)h(x) increases by 800%800\%\newline(B) h(x)h(x) increases by a factor of 88\newline(C) h(x)h(x) decreases by 88\newline(D) h(x)h(x) increases by x=1x = 100
  1. Calculate h(x)h(x): Calculate h(x)h(x) when x=1x = 1.\newlineh(1)=81=8h(1) = 8^1 = 8.
  2. Calculate h(x)h(x): Calculate h(x)h(x) when x=2x = 2.\newlineh(2)=82=64h(2) = 8^2 = 64.
  3. Find factor increase: Find the factor by which h(x)h(x) increases from x=1x = 1 to x=2x = 2.\newlineFactor increase = h(2)h(1)=648=8\frac{h(2)}{h(1)} = \frac{64}{8} = 8.
  4. Match factor increase: Match the factor increase with the given choices.\newlineThe correct choice is (B) h(x)h(x) increases by a factor of 88.

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