Exponential equations and functions

Recognize the structure. Set t = 0 for the initial value and increase t by 1 to see the monthly multiplier.

Work it through. N(0) = 650(1.04)⁰ = 650. Also,

$\frac{\text{N(t + 1)}}{\text{N(t)}}$ = 1.04.

Thus the modeled amount starts at 650 and increases by 4% each month.

Answer: Initial amount 650; monthly growth factor 1.04, or 4% growth per month.

Check. After one month the model gives 676, which is 26 more than 650, and 26/650 = 0.04.

Avoid the trap. 1.04 is not a 104% increase. It represents the original 100% plus a 4% increase.

Recognize the structure. Losing 12% means retaining 88%, so the yearly multiplier is 0.88.

Work it through. Multiply the initial value by that factor once per year:

V(t) = 18000(0.88)^t.

The model can allow real t ≥ 0 if time is continuous, or integer t if only year-end values are considered.

Answer: V(t) = 18000(0.88)^t.

Check. V(1) = 15840, which is 2160 = 0.12(18000) below the initial value.

Avoid the trap. 18000(0.12)^t would mean retaining only 12% each year, a much larger decline. A fixed subtraction each year would instead be a linear model.

Recognize the structure. Nine hours contain three periods of length three hours.

Work it through. M(9) = 96 ($\frac{\text{1}}{\text{2}}$)^9/3 = 96 ($\frac{\text{1}}{\text{2}}$)³ = $\frac{\text{96}}{\text{8}}$ = 12.

Answer: 12 grams.

Check. Successive three-hour amounts are 96, 48, 24, and 12 grams.

Avoid the trap. The exponent is t/3, not 3t. The factor 1/2 applies every three hours, not every hour.

Recognize the structure. 32 = 2⁵, so both sides can use base 2.

Work it through. 2^x+1 = 2⁵ x + 1 = 5 x = 4.

Answer: 4.

Check. 2⁴⁺¹ = 2⁵ = 32.

Avoid the trap. The exponent is the entire expression x + 1. Matching x directly to 5 ignores its constant term.

Recognize the structure. The inputs are equally spaced, so consecutive output ratios reveal the per-unit factor.

Work it through. Each ratio is 3: 15/5 = 45/15 = 135/45 = 3. Therefore b = 3. At t = 0, f(0) = A = 5.

Answer: f(t) = 5 ⋅ 3^t.

Check. The rule gives 5 ⋅ 3³ = 135 at t = 3 and reproduces every listed value.

Avoid the trap. The first differences are 10, 30, and 90, not a constant. A linear rule is not consistent with this table.

Recognize the structure. The output ratio covers a three-unit input increase, not a one-unit increase.

Work it through. Divide the known values:

b⁵⁻² = $\frac{\text{640}}{\text{80}}$ = 8 b³ = 8 b = 2.

Then 80 = A(2²), so A = 20.

Answer: f(0) = 20.

Check. 20 ⋅ 2² = 80 and 20 ⋅ 2⁵ = 640.

Avoid the trap. Using 8 as the per-unit multiplier would compound the full three-unit factor every unit and fail the data.

Recognize the structure. The 8% increase occurs over three years. The one-year factor is the cube root of 1.08.

Work it through. Rewrite the function as

Q(t) = 800 ((1.08)^1/3)^t.

The annual multiplier is (1.08)^1/3. Subtract 1 and multiply by 100 to convert the factor to a percent.

Answer: 100((1.08)^1/3 − 1)%, approximately 2.60% per year.

Check. Cubing the annual factor returns 1.08, the required three-year multiplier.

Avoid the trap. 8%/3 is only a rough simple-rate approximation. It does not reproduce the compound factor exactly.

Recognize the structure. 80 is 1/8 of 640, which is three factors of 1/2.

Work it through. Set the target and divide by 640:

($\frac{\text{1}}{\text{2}}$)^t/6 = $\frac{\text{80}}{\text{640}}$ = $\frac{\text{1}}{\text{8}}$ = ($\frac{\text{1}}{\text{2}}$)³.

Thus t/6 = 3, so t = 18.

Answer: 18 hours.

Check. Three half-lives give $640\to 320\to 160\to 80$, taking 3 ⋅ 6 = 18 hours.

Avoid the trap. The number 3 is a count of half-life periods, not the answer in hours. Convert the count back to the stated time unit.

Recognize the structure. One year contains twelve applications of the factor 1.02.

Work it through. Use the power rule:

P(t) = 1500 ((1.02)¹²)^t.

The annual factor is (1.02)¹² ≈ 1.26824, so the annual percent increase is approximately 100(1.26824 − 1) = 26.8%.

Answer: Annual factor (1.02)¹²; approximately 26.8% annual growth.

Check. Setting t = 1 multiplies 1500 by exactly that annual factor.

Avoid the trap. Twelve times 2% gives 24%, not the compounded annual change. Each month’s increase also grows during later months.

Recognize the structure. Set the exponent equal to zero for the anchor; increase the exponent by 1 for one tripling.

Work it through. At t = 4, the exponent is 0, so N(4) = 200. Increasing t by 5 increases (t − 4)/5 by 1, multiplying the output by 3.

Answer: The modeled amount is 200 at day 4, and it triples every 5 days.

Check. N(9) = 200 ⋅ 3 = 600, while N(4) = 200.

Avoid the trap. The coefficient is not the value at day 0 here. The initial value is 200 ⋅ 3^−4/5 because the exponent has a shift.

Recognize the structure. A ratio removes a and determines b².

Work it through. $\frac{{\text{ab}}^{\text{4}}}{{\text{ab}}^{\text{2}}}$ = $\frac{\text{162}}{\text{18}}$ = 9 b² = 9.

Since b > 0, b = 3. Now ab² = 18 gives 9a = 18, so a = 2.

Answer: 2.

Another route. The midpoint value is f(2) = ab², and f(4) = a(b²)². Thus a = f(2)²/f(4) = 18²/162 = 2.

Avoid the trap. The condition b > 0 is part of the real exponential model. A negative base is not interchangeable with a positive base for arbitrary real inputs.

Recognize the structure. The given four-unit ratio is b⁴ = 5. Express the ten-unit ratio as a power of it.

Work it through. $\frac{\text{g(t + 10)}}{\text{g(t)}}$ = b¹⁰ = (b⁴)^10/4 = 5^5/2 = 25$\sqrt{\text{5}}$.

Both the initial coefficient and the specific starting time cancel.

Answer: 25$\sqrt{\text{5}}$.

Check. Ten units equal 2.5 four-unit periods, so the ratio is 5^2.5. This is greater than 25 and less than 125, as expected.

Avoid the trap. A ten-unit multiplier is not 5(10/4). Repeated proportional change is represented by a power, not linear scaling of the multiplier.

Recognize the structure. Convert minutes to seconds before matching powers of 2.

Work it through. Because t = s/60,

4^t/6 = 4^s/360 = (2²)^s/360 = 2^s/180.

Matching the requested exponent gives k = 180.

Answer: 180.

Check. The original model quadruples every 6 minutes. That is two doublings, so each doubling takes 3 minutes, or 180 seconds.

Avoid the trap. Changing only the time unit would leave base 4. Changing only the base would leave minutes. Both conversions must be reflected in the exponent.

Recognize the structure. The fixed offset cancels in differences, even though it prevents constant ratios of the total outputs.

Work it through. The first differences are −4, −2, and −1. In this model consecutive differences have ratio b, so b = (−2)/(−4) = 1/2. From f(0) = 14 and f(1) = 10,

c + a = 14, c + $\frac{\text{a}}{\text{2}}$ = 10.

Subtracting gives a/2 = 4, so a = 8 and c = 6.

Answer: 6.

Check. 6 + 8(1/2)² = 8 and 6 + 8(1/2)³ = 7, matching the remaining data.

Avoid the trap. The total outputs do not halve. Their excess above 6 halves: 8, 4, 2, 1.

Recognize the structure. First identify the model. Then check neighboring integer inputs rather than rounding a crossing time.

Work it through. The ratio gives b² = 48/12 = 4, so b = 2. From Ab = 12, A = 6. Therefore f(n) = 6 ⋅ 2ⁿ.

Evaluate consecutive powers near the threshold:

f(5) = 6 ⋅ 32 = 192 < 200, f(6) = 6 ⋅ 64 = 384 > 200.

Since the function is strictly increasing, all smaller nonnegative integers give values at most 192.

Answer: 6.

Check. The derived rule gives f(1) = 12 and f(3) = 48, so it satisfies both starting conditions.

Avoid the trap. The answer must be an integer that meets the strict inequality. Rounding a continuous crossing downward would choose a value that has not reached the threshold.