Growth Factor Quiz: Interpreting Initial Value and Growth Factor

A 5% hourly increase means multiplying by 1.05 each hour: N(t) = 1000 times (1.05)^t. Option B adds 0.05 units per hour linearly, which is not 5% growth. Option C uses 0.95, which represents 5% decay not growth. Option D uses 1.5, representing 50% growth per hour, far exceeding 5%.

Divide both sides by a: y/a = b^t. Take the natural logarithm of both sides: ln(y/a) = t times ln(b). Divide by ln(b): t = ln(y/a) divided by ln(b). Option A inverts the argument inside the logarithm. Option B omits the division by a before taking the log. Option D inverts the entire fraction. Only option C correctly isolates t.

N(2) = 120 times (1.5)^2 = 120 times 2.25 = 270. Option B gives 180 = 120 times 1.5, which is N(1) not N(2). Option C gives 240 = 120 times 2, incorrectly using 2 instead of 2.25. Option D gives 300 = 120 times 2.5, using an incorrect squared value. Only 270 correctly applies (1.5)^2 = 2.25.

Growth requires b > 1 and a > 0. Option A has a = 50 > 0 and b = 1.1 > 1, confirming growth. Option D has a = 3 > 0 and b = 1.5 > 1, also confirming growth. Option B has a = 0, making the function identically 0 for all t — a degenerate case. Option C has b = 0.9 < 1, representing exponential decay not growth.

After 2 periods: y(2) = a times (0.8)^2 = a times 0.64 = 0.64a. Option B gives 0.80a, which is only after 1 period. Option C gives 0.36a = a times 0.6^2, using the wrong base. Option D gives 1.60a suggesting growth, but b=0.8 < 1 means decay. Only 0.64a correctly squares the base 0.8.

The answer is True. Increasing t by 3 gives N(t+3) = a times b^((t+3)/3) = a times b^(t/3) times b^(3/3) = a times b^(t/3) times b. The quantity is multiplied by b every 3 time units. This is the defining property of the model — b is the per-3-unit growth factor, not the per-unit factor.

Percent change = (b - 1) times 100 percent. For b > 1 this gives positive growth — for example b = 1.25 gives 25% growth. For 0 < b < 1 it gives negative decay — for example b = 0.9 gives -10% decay. Option B gives 100b, which would give 125% for b=1.25 instead of 25%. Option C omits the multiplication by 100. Option D has no valid derivation.

P(0) = a = 50. Then b = P(1) divided by a = 60 divided by 50 = 1.2. The pair is (50, 1.2). Option A uses 60 as the initial value, ignoring P(0)=50. Option C gives b=1.1, which would produce P(1) = 50 times 1.1 = 55, not 60. Option D has both values wrong.

P(3) = a times b^3 = 2a, so b^3 = 2. Taking the cube root gives b = 2^(1/3) approximately 1.26. Option B gives 1/2, which would give b^3 = 1/8, not 2. Option C gives 3/2, where (3/2)^3 = 27/8 not 2. Option D gives sqrt(2), where (sqrt(2))^3 = 2sqrt(2) not 2. Only 2^(1/3) satisfies b^3 = 2.

Since b is between 0 and 1, repeated multiplication shrinks the quantity confirming exponential decay, confirming A. Percent change = (b-1) times 100 is negative when b < 1, confirming B. A half-life exists because there is always a time T where b^T = 1/2, confirming C. Option D is false — with a > 0 and b > 0, every power of b is positive, so y(t) remains positive for all t and never becomes negative.

In N(t) = a times b^t, evaluating at t=0 gives N(0) = a times b^0 = a times 1 = a. Here N(0) = 300, so 300 is the initial amount. Option B describes 1.25, the base. Option C describes the percent rate derived from the base. Option D gives the amount after one period, which would be 300 times 1.25 = 375, not 300.

The answer is True. With b = 1, N(t) = a times 1^t = a times 1 = a for all values of t. Every power of 1 equals 1, so the function never changes regardless of t. This represents a constant function with zero percent change per period, fitting neither growth nor decay.

At t = 0, S(0) = 800 times (0.75)^0 = 800 times 1 = 800. The initial value of any exponential function a times b^t is always a when t = 0, regardless of the base. Option A gives 600 = 800 times 0.75, which is S(1) not S(0). Options B and C have no direct relationship to the given function.

Doubling time T satisfies a times b^T = 2a, so b^T = 2. Taking natural logarithm: T times ln(b) = ln(2), giving T = ln(2) divided by ln(b) = ln(2) divided by ln(1.2). Option B divides 2 by 1.2 with no logarithmic basis. Option C inverts the numerator and denominator. Option D gives 0.5, which has no relationship to the base 1.2.

P(0) = a = 200. Then P(1) = a times b = 260, so b = 260 divided by 200 = 1.3. The pair is (200, 1.3). Option B uses 260 as the initial value, ignoring that P(0)=200. Option C gives b=1.26, which would produce P(1)=200 times 1.26=252, not 260. Option D gives b=0.77, producing decay rather than growth.

The initial value a = 500 is confirmed by M(0) = 500 times 1 = 500, confirming A. The base b = 1.03 is the per-period multiplier, confirming B. The percent increase per period = (1.03-1) times 100 = 3%, confirming D. Option C is false — the function multiplies by 1.03 each period, it does not add a fixed 3 units. The actual amount added varies as the quantity grows.

Percent change = (b - 1) times 100 = (0.9 - 1) times 100 = -10%. The negative sign confirms decay. Option A gives -5%, corresponding to b = 0.95. Option C gives -15%, corresponding to b = 0.85. Option D gives -20%, corresponding to b = 0.80. Only -10% correctly applies the formula with b = 0.9.

The answer is True. Evaluating at t = 0 gives N(0) = a times b^0 = a times 1 = a. The coefficient a is always the initial value regardless of the base b. This is why a is called the initial value or starting amount in exponential models.

Percent rate per period = (b - 1) times 100 = (1.25 - 1) times 100 = 25%. Option B gives 1.25%, which would correspond to b = 1.0125. Option C gives -25%, which would require b = 0.75 indicating decay. Option D gives 75%, which would correspond to b = 1.75. Only 25% correctly applies the formula with b = 1.25.

The base b = 1.25 is the multiplicative growth factor applied each period. Each time t increases by 1, N is multiplied by 1.25. Option A describes 300, the initial value. Option C is incorrect because b > 1 means growth, not decay. Option D gives the percent rate 25%, not the factor 1.25 itself — these are related but distinct quantities.