SAT Exponential Functions Mastery Practice Quiz

We have \( 600(0.1)^x = 0.6 \). Dividing both sides by 600 gives \( (0.1)^x = 0.001 \). Recognizing that \( 0.001 = (0.1)^3 \), we can equate exponents: \( x = 3 \).

Starting with \( 400(0.2)^x = 10 \), divide both sides by 400 to get \( (0.2)^x = 0.025 \). Taking the logarithm of both sides, we get \( x \log(0.2) = \log(0.025) \). Solving for \( x \) yields \( x = \\frac{\log(0.025)}{\log(0.2)} \approx 2.3219 \).

We have \( 300(0.3)^x = 30 \). Dividing both sides by 300 gives \( (0.3)^x = 0.1 \). Taking the logarithm of both sides, we get \( x \log(0.3) = \log(0.1) \). Solving for \( x \) yields \( x = \\frac{\log(0.1)}{\log(0.3)} \approx 2.0959 \).

Start with \( 200(0.5)^x = 50 \). Divide both sides by 200 to get \( (0.5)^x = 0.25 \). Recognizing that \( 0.25 = (0.5)^2 \), we can equate exponents: \( x = 2 \).

We start with \( 500(0.4)^a = 100 \). Dividing both sides by 500 gives \( (0.4)^a = 0.2 \). To solve for \( a \), we take the logarithm of both sides: \( a \log(0.4) = \log(0.2) \). Solving for \( a \) yields \( a = \\frac{\log(0.2)}{\log(0.4)} \approx 1.585 \).

Substitute \( 0 \) for \( x \): \( s(0) = 1500(0.9)^0 \). Since \( (0.9)^0 = 1 \), the result is \( 1500 \times 1 = 1500 \).

Substitute \( -1 \) for \( x \): \( r(-1) = 1200(0.15)^{-1} \). Since \( (0.15)^{-1} \approx 6.67 \), the result is approximately \( 1200 \times 6.67 = 8000 \).

Substitute \( 0 \) for \( x \): \( q(0) = 1000(0.8)^0 \). Since \( (0.8)^0 = 1 \), the result is \( 1000 \times 1 = 1000 \).

Substitute \( 1 \) for \( x \): \( p(1) = 900(0.6)^1 \). This simplifies to \( 900 \times 0.6 = 540 \).

Substitute \( 0 \) for \( x \): \( n(0) = 800(0.7)^0 \). Since \( (0.7)^0 = 1 \), the result is \( 800 \times 1 = 800 \).

Substitute \( -2 \) for \( x \): \( m(-2) = 700(0.25)^{-2} \). Since \( (0.25)^{-2} = 16 \), the expression becomes \( 700 \times 16 = 11200 \).

Substitute \( 0 \) for \( x \): \( k(0) = 600(0.1)^0 \). Since \( (0.1)^0 = 1 \), the result is \( 600 \times 1 = 600 \).

Substitute \( 2 \) for \( x \): \( j(2) = 400(0.3)^2 \). This simplifies to \( 400 \times 0.09 = 36 \).

Substitute \( -1 \) for \( x \): \( h(-1) = 200(0.5)^{-1} \). Since \( (0.5)^{-1} = 2 \), the expression becomes \( 200 \times 2 = 400 \).

To find \( g(1) \), substitute \( 1 \) for \( x \) in the function: \( g(1) = 500(0.4)^1 \). This simplifies to \( 500 \times 0.4 = 200 \).