SAT Exponential Functions with Positive and Negative Exponents Practice Quiz

First, solve for \( a \) in \( 400(1.2)^a = 600 \). Dividing both sides by 400 gives \( (1.2)^a = 1.5 \). Taking the natural logarithm of both sides: \( \\ln((1.2)^a) = \\ln(1.5) \). Using the property of logarithms, we get \( a \\ln(1.2) = \\ln(1.5) \). Solving for \( a \) gives \( a = \\frac{\\ln(1.5)}{\\ln(1.2)} \approx 2.127 \). Next, find \( m(a+1) = 400(1.2)^{a+1} = 400(1.2)^{2.127+1} = 400(1.2)^{3.127} \approx 900 \).

We need to solve \( 300(0.7)^x = 210 \). Dividing both sides by 300 gives \( (0.7)^x = 0.7 \). Taking the natural logarithm of both sides: \( \\ln((0.7)^x) = \\ln(0.7) \). Using the property of logarithms, we get \( x \\ln(0.7) = \\ln(0.7) \). Solving for \( x \) gives \( x = \\frac{\\ln(0.7)}{\\ln(0.7)} = 1 \).

We need to solve \( 200(0.8)^x = 100 \). Dividing both sides by 200 gives \( (0.8)^x = 0.5 \). Taking the natural logarithm of both sides: \( \\ln((0.8)^x) = \\ln(0.5) \). Using the property of logarithms, we get \( x \\ln(0.8) = \\ln(0.5) \). Solving for \( x \) gives \( x = \\frac{\\ln(0.5)}{\\ln(0.8)} \approx 3.106 \).

We need to solve \( 50(0.6)^a = 150 \). Dividing both sides by 50 gives \( (0.6)^a = 3 \). Taking the natural logarithm of both sides: \( \\ln((0.6)^a) = \\ln(3) \). Using the property of logarithms, we get \( a \\ln(0.6) = \\ln(3) \). Solving for \( a \) gives \( a = \\frac{\\ln(3)}{\\ln(0.6)} \approx -2.045 \).

First, find \( f(2) \): \( f(2) = 100(1.5)^2 = 100 \\times 2.25 = 225 \). Next, find \( f(-1) \): \( f(-1) = 100(1.5)^{-1} = 100 \\times \\frac{1}{1.5} = 100 \\times \\frac{2}{3} = \\frac{200}{3} \\approx 66.67 \). Adding these values together gives \( 225 + 66.67 = 291.67 \).

To find \( t(1) \), substitute \( 1 \) for \( x \): \( t(1) = 600(0.85)^1 \). Since \( (0.85)^1 = 0.85 \), the expression becomes \( 600 \\times 0.85 = 510 \).

To find \( s(0) \), substitute \( 0 \) for \( x \): \( s(0) = 1000(0.2)^0 \). Since \( (0.2)^0 = 1 \), the expression becomes \( 1000 \\times 1 = 1000 \).

To find \( r(3) \), substitute \( 3 \) for \( x \): \( r(3) = 800(1.1)^3 \). Since \( (1.1)^3 = 1.331 \), the expression becomes \( 800 \\times 1.331 = 1064.8 \).

To find \( q(2) \), substitute \( 2 \) for \( x \): \( q(2) = 250(0.5)^2 \). Since \( (0.5)^2 = 0.25 \), the expression becomes \( 250 \\times 0.25 = 62.5 \).

To find \( p(-1) \), substitute \( -1 \) for \( x \): \( p(-1) = 900(0.9)^{-1} \). Since \( (0.9)^{-1} = \\frac{1}{0.9} = 1.1111... \), the expression becomes \( 900 \\times 1.1111... = 1000 \).

To find \( n(-2) \), substitute \( -2 \) for \( x \): \( n(-2) = 400(0.8)^{-2} \). Since \( (0.8)^{-2} = \\left(\\frac{1}{0.8}\\right)^2 = 1.5625 \), the expression becomes \( 400 \\times 1.5625 = 625 \).

To find \( m(3) \), substitute \( 3 \) for \( x \): \( m(3) = 300(0.6)^3 \). Since \( (0.6)^3 = 0.216 \), the expression becomes \( 300 \\times 0.216 = 64.8 \).

To find \( k(1) \), substitute \( 1 \) for \( x \): \( k(1) = 750(0.75)^1 \). Since \( (0.75)^1 = 0.75 \), the expression becomes \( 750 \\times 0.75 = 562.5 \).

To find \( h(2) \), substitute \( 2 \) for \( x \): \( h(2) = 200(1.2)^2 \). Since \( (1.2)^2 = 1.44 \), the expression becomes \( 200 \\times 1.44 = 288 \).

To find \( g(-1) \), substitute \( -1 \) for \( x \): \( g(-1) = 500(0.4)^{-1} \). Since \( (0.4)^{-1} = \\frac{1}{0.4} = 2.5 \), the expression becomes \( 500 \\times 2.5 = 1250 \).