SAT Exponential Growth: The "1 ยฑ r" Trick

Don't let "compounded annually" scare you. It's just a single formula you can solve in your head.

The Problem

A population of bacteria starts at 500 and increases by 12% every year. Which equation represents the population after \(t\) years?

A) \(P = 500(0.12)^t\)
B) \(P = 500(1.12)^t\)
C) \(P = 500(1.12t)\)

The Trap: Picking 0.12 as the base. If it's growth, the base must be greater than 1. If it's decay, it must be less than 1.

โœ… The Master Formula: \(y = a(1 \pm r)^t\)

  • \(a\) = Initial Amount (The number at the start)
  • \(1 + r\) = Growth (e.g., 12% increase becomes 1.12)
  • \(1 - r\) = Decay (e.g., 12% decrease becomes 0.88)
  • \(t\) = Time (Usually an exponent)

Cheat Sheet for Rates

โ†‘

"Increases by 5%"

Base = \(1 + 0.05 = 1.05\)

โ†“

"Decreases by 5%"

Base = \(1 - 0.05 = 0.95\)

x

"Doubles"

Base = \(2\)

Growth Visualizer

Adjust the rate to see how the formula and graph change instantly.

Your Equation:
y = 500(1.12)แต—

Exponential Growth ๐Ÿ“ˆ

๐ŸŽ“ The Proof: Why $y = a(1 \pm r)^t$?

Why do we add 1 to the rate? The derivation reveals that exponential growth is actually just repetitive factoring.

1. Start with an initial amount $a$ and a growth rate $r$. After the first time period, the new amount is the original plus the increase:

$Amount_1 = a + (r \cdot a)$

2. Factor out the $a$ to see the Multiplier:

$Amount_1 = a(1 + r)$

3. To find the amount after the second period, apply the same multiplier to the new amount:

$Amount_2 = [a(1 + r)] \cdot (1 + r) = a(1 + r)^2$

4. The Pattern: Each time period $t$ simply adds another power to the multiplier:

$y = a(1 + r)^t$

This explains why the base is $1.12$ for a $12\%$ increase. The "1" represents the 100% you already had, and the "0.12" represents the new growth.

Compounding is the 8th wonder of the world. โ€” Practix Team

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