Exponential Growth and Decay: Practice

Losing $15\%$ of the value means keeping the other $85\%$, so one year multiplies the value by $1 - 0.15$.

$b = 1 - 0.15 = 0.85$

The rate here is $r = -0.15$, but the factor you multiply by is $0.85$.

The factor is $b = 1 + r$, so the rate is recovered by subtracting $1$.

$r = b - 1 = 1.09 - 1 = 0.09$

A decimal rate of $0.09$ is $9\%$ per step.

Subtract $1$ from the factor to get the rate.

$r = 0.72 - 1 = -0.28$

A rate of $-0.28$ is a $28\%$ decrease. The number $72\%$ is what is left behind, not what is lost.

With a positive starting amount, the model decays exactly when the factor sits strictly between $0$ and $1$, since each step then multiplies by less than one.

$0 < 0.8 < 1$

The factors $1.2$ and $2$ are greater than $1$, so those models grow, and a factor of $1$ leaves the amount unchanged forever.

The starting amount is the coefficient and the growth factor is the base. A rate of $5\%$ means $r = 0.05$, so the factor is

$b = 1 + 0.05 = 1.05$

which gives $A = 1200(1.05)^t$. Using $0.05$ as the base would destroy $95\%$ of the money every year.

Doubling is a growth factor of $2$, and the starting amount $7$ is the coefficient that multiplies the power.

$A = 7(2)^t$

Check it at $t = 0$: the amount is $7 \cdot 2^0 = 7$, the starting value. Swapping the $7$ and the $2$ would start the model at $2$ instead.

Tripling is a growth factor of $3$, applied once each hour, so two hours multiply by $3^2$.

$200(3)^2 = 200 \cdot 9 = 1800$

The count runs $200 \to 600 \to 1800$. Stopping at $600$ counts only one hour.

Write the percent as a decimal first: $3.5\% = 0.035$. Then add $1$.

$b = 1 + 0.035 = 1.035$

The factor $1.35$ would be a $35\%$ increase, ten times too large.

Subtract $1$ from the factor to read off the rate.

$r = 0.94 - 1 = -0.06$

The negative sign says the quantity falls, and the size $0.06$ says it falls by $6\%$ each step. The $94\%$ is the portion that survives.

The growth factor is $1 + 0.04 = 1.04$, applied once for each of the two years.

$400(1.04)^2 = 400(1.0816) = 432.64$

The answer $432$ dollars comes from adding $16$ dollars twice, which forgets that the second year's interest is earned on $416$ dollars, not on $400$.

First count the half-lives: $18$ days divided by a $6$-day half-life is $3$ of them.

$64\left(\tfrac{1}{2}\right)^{18/6} = 64\left(\tfrac{1}{2}\right)^{3} = \frac{64}{8} = 8$

The sample runs $64 \to 32 \to 16 \to 8$ grams.

Each step multiplies the current amount by $b$, so the amount rises exactly when that multiplier is bigger than $1$.

$b > 1 \quad \Longrightarrow \quad P\,b^{\,t+1} > P\,b^{\,t}$

A factor between $0$ and $1$ shrinks the amount, and $b = 1$ leaves it fixed.