Exponential Functions worksheet with 14 problems for evaluating functions at specific values.

Let’s solve each problem one by one. We’ll plug in the given value for the variable and calculate step by step. Remember: when you see a negative exponent, it means “take the reciprocal” — like \( a^{-n} = \frac{1}{a^n} \). And we’ll round to the nearest hundredth if needed.

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1) \( g(n) = 3 \cdot \left(\frac{5}{6}\right)^n \) at \( n = -3 \)

→ \( g(-3) = 3 \cdot \left(\frac{5}{6}\right)^{-3} = 3 \cdot \left(\frac{6}{5}\right)^3 \)

→ \( \left(\frac{6}{5}\right)^3 = \frac{216}{125} = 1.728 \)

→ \( 3 \cdot 1.728 = 5.184 \) → Round to nearest hundredth: 5.18

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2) \( h(x) = 9 \cdot \left(\frac{1}{2}\right)^x \) at \( x = 3 \)

→ \( h(3) = 9 \cdot \left(\frac{1}{2}\right)^3 = 9 \cdot \frac{1}{8} = \frac{9}{8} = 1.125 \) → 1.13

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3) \( f(n) = \frac{4}{7} \cdot \left(\frac{1}{2}\right)^n \) at \( n = -3 \)

→ \( f(-3) = \frac{4}{7} \cdot \left(\frac{1}{2}\right)^{-3} = \frac{4}{7} \cdot 2^3 = \frac{4}{7} \cdot 8 = \frac{32}{7} ≈ 4.571... \) → 4.57

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4) \( h(n) = \frac{1}{7} \cdot 2^n \) at \( n = 2 \)

→ \( h(2) = \frac{1}{7} \cdot 2^2 = \frac{1}{7} \cdot 4 = \frac{4}{7} ≈ 0.571... \) → 0.57

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5) \( g(y) = \frac{9}{3} \cdot \left(\frac{1}{2}\right)^y \) at \( y = 2 \)

→ First, simplify \( \frac{9}{3} = 3 \)

→ \( g(2) = 3 \cdot \left(\frac{1}{2}\right)^2 = 3 \cdot \frac{1}{4} = \frac{3}{4} = 0.75 \) → 0.75

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6) \( h(x) = 5 \cdot 2^x \) at \( x = 3 \)

→ \( h(3) = 5 \cdot 2^3 = 5 \cdot 8 = 40 \) → 40.00

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7) \( f(x) = 3 \cdot \left(\frac{5}{7}\right)^x \) at \( x = 3 \)

→ \( f(3) = 3 \cdot \left(\frac{5}{7}\right)^3 = 3 \cdot \frac{125}{343} = \frac{375}{343} ≈ 1.093... \) → 1.09

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8) \( h(n) = \frac{1}{2} \cdot \left(\frac{1}{3}\right)^n \) at \( n = -2 \)

→ \( h(-2) = \frac{1}{2} \cdot \left(\frac{1}{3}\right)^{-2} = \frac{1}{2} \cdot 3^2 = \frac{1}{2} \cdot 9 = 4.5 \) → 4.50

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9) \( g(y) = \frac{2}{7} \cdot 2^y \) at \( y = 2 \)

→ \( g(2) = \frac{2}{7} \cdot 2^2 = \frac{2}{7} \cdot 4 = \frac{8}{7} ≈ 1.142... \) → 1.14

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10) \( f(y) = \frac{3}{2} \cdot 2^y \) at \( y = -2 \)

→ \( f(-2) = \frac{3}{2} \cdot 2^{-2} = \frac{3}{2} \cdot \frac{1}{4} = \frac{3}{8} = 0.375 \) → 0.38

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11) \( f(x) = 4 \cdot 2^x \) at \( x = -2 \)

→ \( f(-2) = 4 \cdot 2^{-2} = 4 \cdot \frac{1}{4} = 1 \) → 1.00

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12) \( h(y) = \frac{1}{2} \cdot \left(\frac{1}{3}\right)^y \) at \( y = 3 \)

→ \( h(3) = \frac{1}{2} \cdot \left(\frac{1}{3}\right)^3 = \frac{1}{2} \cdot \frac{1}{27} = \frac{1}{54} ≈ 0.0185... \) → 0.02

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13) \( g(y) = \frac{1}{2} \cdot \left(\frac{2}{3}\right)^y \) at \( y = -2 \)

→ \( g(-2) = \frac{1}{2} \cdot \left(\frac{2}{3}\right)^{-2} = \frac{1}{2} \cdot \left(\frac{3}{2}\right)^2 = \frac{1}{2} \cdot \frac{9}{4} = \frac{9}{8} = 1.125 \) → 1.13

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14) \( g(y) = 8 \cdot \left(\frac{1}{2}\right)^y \) at \( y = -2 \)

→ \( g(-2) = 8 \cdot \left(\frac{1}{2}\right)^{-2} = 8 \cdot 2^2 = 8 \cdot 4 = 32 \) → 32.00

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Final Answer:
1) 5.18
2) 1.13
3) 4.57
4) 0.57
5) 0.75
6) 40.00
7) 1.09
8) 4.50
9) 1.14
10) 0.38
11) 1.00
12) 0.02
13) 1.13
14) 32.00