Exponential Relationships

Exponential growth multiplies by a constant ratio greater than $1$. Here each term is $2$ times the previous: $2, 4, 8, 16, 32$. ($2, 4, 6, 8, 10$ has a constant difference (linear); $1, 4, 9, 16, 25$ are squares (quadratic).)

In $15$ years there are $\dfrac{15}{5} = 3$ half-lives, so the sample halves three times: $240 \times \left(\dfrac{1}{2}\right)^3 = 240 \times \dfrac{1}{8} = 30$ g. (This is exponential decay with base $\tfrac{1}{2}$.)

An exponential has the form $y = ab^x$ with the variable in the exponent. $y = 5 \times 2^x$ matches. ($y = x^2$ is quadratic; $y = 2x + 3$ is linear; $y = \dfrac{1}{x}$ is reciprocal.)

$2^{x + 2} = 2^x \times 2^2 = 32 \times 4 = 128$ (no need to find $x$ first). ($34$ added $2$ to 32; $64$ only multiplied by $2$ once.)

Multiply the two factors: $1.20 \times 0.80 = 0.96$, which is $96\%$ of the original, i.e. $4\%$ less. (The percentages do not cancel, because the $20\%$ decrease acts on the larger, increased price.)

A negative exponent gives a reciprocal: $2^{-3} = \dfrac{1}{2^3} = \dfrac{1}{8}$, always positive. ($-8$ confused $2^{-3}$ with $-2^3$; the value of $2^x$ is never negative.)

For $0 < b < 1$, $b^x$ keeps halving (or shrinking) toward 0 but stays positive, so $y$ never actually reaches 0. The $x$-axis is a horizontal asymptote. (This "approaches but never reaches" behaviour is the key feature of exponential decay.)

$0.5^3 = 0.125$, then $100 \times 0.125 = 12.5$ (apply the exponent first). ($50$ used only one factor of $0.5$.) The value halves each step: $100 \to 50 \to 25 \to 12.5$.

Divide first: $2^x = \dfrac{96}{3} = 32$. Then $32 = 2^5$, so $x = 5$. ($32$ stopped at $2^x$ without finding the index.)

Each bounce multiplies the height by $0.8$, so after two bounces the height is $5 \times 0.8^2 = 5 \times 0.64 = 3.2$ m. ($4$ m is only the first bounce.)

Each year multiplies by $1.08$, so three years multiply by $1.08 \times 1.08 \times 1.08 = 1.08^3$. ($1.24$ is the simple-interest factor $1 + 0.08 \times 3$; $0.92^3$ would be an $8\%$ decay.)

$A = 1000 \times (1.10)^2 = 1000 \times 1.21 = \$1210$. ($\$1200$ used simple interest; $\$1331$ used $3$ years.)

Linear growth adds a constant amount, but exponential growth adds a constant percentage, so each year's increase itself keeps getting bigger. Given enough time, exponential growth always overtakes linear growth, so account B becomes larger. (Account A may lead early on, but B eventually wins.)

$A = 500 \times (1.04)^3 = 500 \times 1.124864 \approx \$562.43$. ($\$560$ used simple interest.)

$1000(1 + r)^2 = 1210$, so $(1 + r)^2 = 1.21$ and $1 + r = \sqrt{1.21} = 1.1$, giving $r = 0.1 = 10\%$. ($21\%$ mistook the total $21\%$ growth for the annual rate.)

At $x = 0$: $y = 3 \times 2^0 = 3 \times 1 = 3$. In $y = ab^x$ the y-intercept is always $a$. ($2$ gave the base.)

Write both sides with base $2$: $8^x = 2^{3x}$, so $2^{x + 3} = 2^{3x}$. Equating the indices, $x + 3 = 3x$, so $2x = 3$ and $x = \dfrac{3}{2}$.

Each fold doubles the thickness, so after $10$ folds it is $0.1 \times 2^{10} = 0.1 \times 1024 = 102.4$ mm. (Doubling ten times is exponential growth, not adding $10$.)

After $n$ hours there are $100 \times 2^n$ bacteria. We need $2^n > 10$: $2^3 = 8$ gives $800$ (not enough) and $2^4 = 16$ gives $1600$ (enough). So the colony first exceeds $1000$ after $4$ hours.

$y = 2 \times 3^x$ gives $y(0) = 2 \times 3^0 = 2$ and grows because the base $3 > 1$. ($y = 2 \times 0.5^x$ passes through $(0, 2)$ but decays; $y = 2x$ is linear.)

Write both sides with base $2$: $8^x = 2^{3x}$ and $4 = 2^2$, so $3x = 2$ and $x = \dfrac{2}{3}$. ($2$ solved $2^x = 4$ instead.)

The curve doubles with each unit step to the right, reaching $16$ at $x = 4$ and $32$ at $x = 5$. Starting from $(0, 1)$, this is $y = 2^x$. ($y = 2 \times 2^x$ would already reach $32$ at $x = 4$; $y = 3^x$ climbs far more steeply; $y = 0.5^x$ decays and $y = x^2$ is a parabola.)

Each year it keeps $85\% = 0.85$: $20{,}000 \times 0.85^2 = 20{,}000 \times 0.7225 = \$14{,}450$. ($\$17{,}000$ used only one year.)

The $y$-intercept is $100$, so $a = 100$. The value halves each unit step ($100 \to 50 \to 25$), so the base is $0.5$, giving $y = 100 \times 0.5^x$. ($y = 100 \times 0.25^x$ would fall to $25$ by $x = 1$; $y = 100 - 25x$ is a straight line; $y = 100 \times 2^x$ grows.)

Number of doublings: $\dfrac{12}{3} = 4$. So $50 \times 2^4 = 50 \times 16 = 800$ cells. ($400$ doubled three times only.)