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Monday, August 31, 2026

SAT Exponential Functions: Growth, Decay, Formulas & Easy Tricks


๐™Ž๐˜ผ๐™ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ๐™จ ๐™‚๐™ช๐™ž๐™™๐™š

๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™, ๐˜ฟ๐™š๐™˜๐™–๐™ฎ, ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š๐™จ, ๐™‚๐™ง๐™–๐™ฅ๐™๐™จ, ๐™๐™–๐™—๐™ก๐™š๐™จ ๐™–๐™ฃ๐™™ ๐™’๐™ค๐™ง๐™™ ๐™‹๐™ง๐™ค๐™—๐™ก๐™š๐™ข๐™จ

An exponential function can look complicated at first.

But underneath the equation, table, graph or word problem, there is usually one simple idea:

๐™๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™ข๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ž๐™š๐™ง ๐™ž๐™จ ๐™–๐™ฅ๐™ฅ๐™ก๐™ž๐™š๐™™ ๐™–๐™œ๐™–๐™ž๐™ฃ ๐™–๐™ฃ๐™™ ๐™–๐™œ๐™–๐™ž๐™ฃ.

That single idea connects exponential equations, exponential growth, exponential decay, percentage changes, doubling, halving, tables and graphs.

This guide brings those ideas together in one place.


✦ ๐Ÿญ. ๐™’๐™๐™–๐™ฉ ๐™„๐™จ ๐˜ผ๐™ฃ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ?

A common exponential function is written as:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

There are three important parts.

๐’‚ = starting value

๐’ƒ = multiplication factor

๐’™ = number of repeated changes

The most important clue is that the variable appears in the exponent.

For example:

๐’‡(๐’™) = ๐Ÿฑ(๐Ÿฎหฃ)

is exponential because ๐’™ is in the exponent.

But:

๐’‡(๐’™) = ๐Ÿฑ๐’™²

is not an exponential function.

Here, the variable is the base and the exponent is fixed.


✦ ๐Ÿฎ. ๐™๐™๐™š ๐™ˆ๐™–๐™ž๐™ฃ ๐™„๐™™๐™ša: ๐˜ผ๐™™๐™™ ๐™Š๐™ง ๐™ˆ๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ฎ?

This is one of the quickest ways to distinguish linear and exponential patterns.

Consider:

๐Ÿฏ, ๐Ÿฒ, ๐Ÿต, ๐Ÿญ๐Ÿฎ, ๐Ÿญ๐Ÿฑ

The same amount is added each time:

+๐Ÿฏ

This is a linear pattern.

Now consider:

๐Ÿฏ, ๐Ÿฒ, ๐Ÿญ๐Ÿฎ, ๐Ÿฎ๐Ÿฐ, ๐Ÿฐ๐Ÿด

Each value is multiplied by:

×๐Ÿฎ

This is an exponential pattern.

๐™๐™š๐™ข๐™š๐™ข๐™—๐™š๐™ง:

๐™Ž๐™–๐™ข๐™š ๐™™๐™ž๐™›๐™›๐™š๐™ง๐™š๐™ฃ๐™˜๐™š → ๐™ก๐™ž๐™ฃ๐™š๐™–๐™ง

๐™Ž๐™–๐™ข๐™š ๐™ง๐™–๐™ฉ๐™ž๐™ค → ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก


✦ ๐Ÿฏ. ๐™’๐™๐™–๐™ฉ ๐˜ฟ๐™ค๐™š๐™จ ๐’‚ ๐™ˆ๐™š๐™–๐™ฃ?

Look at:

๐’‡(๐’™) = ๐Ÿด(๐Ÿฏหฃ)

The starting value is:

๐’‚ = ๐Ÿด

Why?

Set:

๐’™ = ๐Ÿฌ

Then:

๐’‡(๐Ÿฌ) = ๐Ÿด(๐Ÿฏ⁰)

Since:

๐Ÿฏ⁰ = ๐Ÿญ

we get:

๐’‡(๐Ÿฌ) = ๐Ÿด

So in:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

the value of ๐’‚ is the output when ๐’™ = ๐Ÿฌ.


✦ ๐Ÿฐ. ๐™’๐™๐™–๐™ฉ ๐˜ฟ๐™ค๐™š๐™จ ๐’ƒ ๐™ˆ๐™š๐™–๐™ฃ?

In:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

the number ๐’ƒ tells you how the output changes when ๐’™ increases by 1.

For example:

๐’‡(๐’™) = ๐Ÿฑ(๐Ÿฎหฃ)

Values include:

๐’‡(๐Ÿฌ) = ๐Ÿฑ

๐’‡(๐Ÿญ) = ๐Ÿญ๐Ÿฌ

๐’‡(๐Ÿฎ) = ๐Ÿฎ๐Ÿฌ

๐’‡(๐Ÿฏ) = ๐Ÿฐ๐Ÿฌ

Every step multiplies the previous value by:

×๐Ÿฎ

So the base is the repeated multiplier.


✦ ๐Ÿฑ. ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™‘๐™š๐™ง๐™จ๐™ช๐™จ ๐˜ฟ๐™š๐™˜๐™–๐™ฎ

The base gives you an immediate clue.

๐™„๐™› ๐’ƒ > ๐Ÿญ:

The function shows exponential growth.

Example:

๐’‡(๐’™) = ๐Ÿญ๐Ÿฌ(๐Ÿญ.๐Ÿฎหฃ)


๐™„๐™› ๐Ÿฌ < ๐’ƒ < ๐Ÿญ:

The function shows exponential decay.

Example:

๐’‡(๐’™) = ๐Ÿญ๐Ÿฌ(๐Ÿฌ.๐Ÿดหฃ)

The values get smaller as ๐’™ increases.

๐™Œ๐™ช๐™ž๐™˜๐™  ๐™˜๐™๐™š๐™˜๐™ :

๐’ƒ > ๐Ÿญ → ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™

๐Ÿฌ < ๐’ƒ < ๐Ÿญ → ๐˜ฟ๐™š๐™˜๐™–๐™ฎ


✦ ๐Ÿฒ. ๐™๐™ช๐™ง๐™ฃ๐™ž๐™ฃ๐™œ ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š๐™จ ๐™„๐™ฃ๐™ฉ๐™ค ๐™ˆ๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ž๐™š๐™ง๐™จ

This is one of the most important skills in exponential word problems.

Suppose something increases by ๐Ÿญ๐Ÿฌ%.

A 10% increase means the new amount is:

๐Ÿญ๐Ÿฌ๐Ÿฌ% + ๐Ÿญ๐Ÿฌ% = ๐Ÿญ๐Ÿญ๐Ÿฌ%

As a decimal:

๐Ÿญ.๐Ÿญ๐Ÿฌ

Therefore:

๐Ÿญ๐Ÿฌ% ๐™ž๐™ฃ๐™˜๐™ง๐™š๐™–๐™จ๐™š → ×๐Ÿญ.๐Ÿญ๐Ÿฌ


Suppose something increases by ๐Ÿฎ๐Ÿฑ%.

๐Ÿญ๐Ÿฌ๐Ÿฌ% + ๐Ÿฎ๐Ÿฑ% = ๐Ÿญ๐Ÿฎ๐Ÿฑ%

Therefore:

๐Ÿฎ๐Ÿฑ% ๐™ž๐™ฃ๐™˜๐™ง๐™š๐™–๐™จ๐™š → ×๐Ÿญ.๐Ÿฎ๐Ÿฑ


✦ ๐Ÿณ. ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š ๐˜ฟ๐™š๐™˜๐™ง๐™š๐™–๐™จ๐™š๐™จ

Suppose something decreases by ๐Ÿญ๐Ÿฌ%.

The amount remaining is:

๐Ÿญ๐Ÿฌ๐Ÿฌ% − ๐Ÿญ๐Ÿฌ% = ๐Ÿต๐Ÿฌ%

As a decimal:

๐Ÿฌ.๐Ÿต๐Ÿฌ

Therefore:

๐Ÿญ๐Ÿฌ% ๐™™๐™š๐™˜๐™ง๐™š๐™–๐™จ๐™š → ×๐Ÿฌ.๐Ÿต๐Ÿฌ

Similarly:

๐Ÿฎ๐Ÿฌ% decrease → ×๐Ÿฌ.๐Ÿด๐Ÿฌ

๐Ÿฏ๐Ÿฌ% decrease → ×๐Ÿฌ.๐Ÿณ๐Ÿฌ

๐Ÿฐ๐Ÿฌ% decrease → ×๐Ÿฌ.๐Ÿฒ๐Ÿฌ

๐Ÿฑ๐Ÿฌ% decrease → ×๐Ÿฌ.๐Ÿฑ๐Ÿฌ


⚠️ ✦ ๐Ÿด. ๐™๐™๐™š ๐˜ฝ๐™ž๐™œ ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š ๐™๐™ง๐™–๐™ฅ

Suppose a quantity decreases by ๐Ÿด๐Ÿฌ%.

The incorrect multiplier is:

๐Ÿฌ.๐Ÿด๐Ÿฌ

Why?

Because 80% is the amount removed, not the amount remaining.

The amount remaining is:

๐Ÿญ๐Ÿฌ๐Ÿฌ% − ๐Ÿด๐Ÿฌ% = ๐Ÿฎ๐Ÿฌ%

Therefore:

๐Ÿด๐Ÿฌ% ๐™™๐™š๐™˜๐™ง๐™š๐™–๐™จ๐™š → ×๐Ÿฌ.๐Ÿฎ๐Ÿฌ

This is an easy place to lose a question.


✦ ๐Ÿต. ๐˜ฝ๐™ช๐™ž๐™ก๐™™๐™ž๐™ฃ๐™œ ๐˜ผ๐™ฃ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™ˆ๐™ค๐™™๐™š๐™ก

Suppose a population begins at:

๐Ÿฎ๐Ÿฌ๐Ÿฌ๐Ÿฌ

and increases by:

๐Ÿฑ% per year

Starting value:

๐’‚ = ๐Ÿฎ๐Ÿฌ๐Ÿฌ๐Ÿฌ

Growth multiplier:

๐Ÿญ + ๐Ÿฌ.๐Ÿฌ๐Ÿฑ = ๐Ÿญ.๐Ÿฌ๐Ÿฑ

Therefore:

๐‘ท(๐’•) = ๐Ÿฎ๐Ÿฌ๐Ÿฌ๐Ÿฌ(๐Ÿญ.๐Ÿฌ๐Ÿฑแต—)

The structure is always:

๐™Ž๐™ฉ๐™–๐™ง๐™ฉ๐™ž๐™ฃ๐™œ ๐™ซ๐™–๐™ก๐™ช๐™š × (๐™œ๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง)แต—


✦ ๐Ÿญ๐Ÿฌ. ๐˜ฟ๐™š๐™˜๐™–๐™ฎ ๐™ˆ๐™ค๐™™๐™š๐™ก๐™จ

Suppose a machine is worth:

$๐Ÿญ๐Ÿฑ๐Ÿฌ๐Ÿฌ๐Ÿฌ

and loses:

๐Ÿญ๐Ÿฎ% of its value each year

The amount remaining each year is:

๐Ÿญ − ๐Ÿฌ.๐Ÿญ๐Ÿฎ = ๐Ÿฌ.๐Ÿด๐Ÿด

Therefore:

๐‘ฝ(๐’•) = ๐Ÿญ๐Ÿฑ๐Ÿฌ๐Ÿฌ๐Ÿฌ(๐Ÿฌ.๐Ÿด๐Ÿดแต—)

Notice something important.

The machine does not lose $1,800 every year.

It loses 12% of its current value.

That distinction creates exponential decay.


✦ ๐Ÿญ๐Ÿญ. ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™‘๐™จ ๐™‡๐™ž๐™ฃ๐™š๐™–๐™ง ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™

Suppose two quantities start at 100.

๐™‡๐™ž๐™ฃ๐™š๐™–๐™ง

Increase by 20 each time:

๐Ÿญ๐Ÿฌ๐Ÿฌ, ๐Ÿญ๐Ÿฎ๐Ÿฌ, ๐Ÿญ๐Ÿฐ๐Ÿฌ, ๐Ÿญ๐Ÿฒ๐Ÿฌ, ๐Ÿญ๐Ÿด๐Ÿฌ

๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก

Increase by 20% each time:

๐Ÿญ๐Ÿฌ๐Ÿฌ, ๐Ÿญ๐Ÿฎ๐Ÿฌ, ๐Ÿญ๐Ÿฐ๐Ÿฐ, ๐Ÿญ๐Ÿณ๐Ÿฎ.๐Ÿด, ๐Ÿฎ๐Ÿฌ๐Ÿณ.๐Ÿฏ๐Ÿฒ

The first adds the same amount.

The second multiplies by the same factor.

๐™๐™๐™–๐™ฉ ๐™ž๐™จ ๐™ฉ๐™๐™š ๐™š๐™จ๐™จ๐™จ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™™๐™ž๐™›๐™›๐™š๐™ง๐™š๐™ฃ๐™˜๐™š.


✦ ๐Ÿญ๐Ÿฎ. ๐™๐™ž๐™ฃ๐™™๐™ž๐™ฃ๐™œ ๐™–๐™ฃ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ ๐™๐™ง๐™ค๐™ข ๐™– ๐™๐™–๐™—๐™ก๐™š

Consider:

๐’™๐’‡(๐’™)
๐Ÿฌ๐Ÿฑ
๐Ÿญ๐Ÿญ๐Ÿฌ
๐Ÿฎ๐Ÿฎ๐Ÿฌ
๐Ÿฏ๐Ÿฐ๐Ÿฌ
๐Ÿฐ๐Ÿด๐Ÿฌ

Look at consecutive ratios.

๐Ÿญ๐Ÿฌ ÷ ๐Ÿฑ = ๐Ÿฎ

๐Ÿฎ๐Ÿฌ ÷ ๐Ÿญ๐Ÿฌ = ๐Ÿฎ

๐Ÿฐ๐Ÿฌ ÷ ๐Ÿฎ๐Ÿฌ = ๐Ÿฎ

๐Ÿด๐Ÿฌ ÷ ๐Ÿฐ๐Ÿฌ = ๐Ÿฎ

The multiplier is:

๐’ƒ = ๐Ÿฎ

The starting value is:

๐’‚ = ๐Ÿฑ

Therefore:

๐’‡(๐’™) = ๐Ÿฑ(๐Ÿฎหฃ)


✦ ๐Ÿญ๐Ÿฏ. ๐™๐™ž๐™ฃ๐™™๐™ž๐™ฃ๐™œ ๐™– ๐™ˆ๐™ž๐™จ๐™จ๐™ž๐™ฃ๐™œ ๐™‘๐™–๐™ก๐™ช๐™š

Suppose:

๐’™๐’‡(๐’™)
๐Ÿฌ๐Ÿฒ
๐Ÿญ๐Ÿญ๐Ÿด
๐Ÿฎ?
๐Ÿฏ๐Ÿญ๐Ÿฒ๐Ÿฎ

The multiplier is:

๐Ÿญ๐Ÿด ÷ ๐Ÿฒ = ๐Ÿฏ

So:

๐Ÿฒ × ๐Ÿฏ = ๐Ÿญ๐Ÿด

๐Ÿญ๐Ÿด × ๐Ÿฏ = ๐Ÿฑ๐Ÿฐ

๐Ÿฑ๐Ÿฐ × ๐Ÿฏ = ๐Ÿญ๐Ÿฒ๐Ÿฎ

Therefore:

๐’‡(๐Ÿฎ) = ๐Ÿฑ๐Ÿฐ

You do not always need to build the entire equation.

Sometimes the pattern is enough.


✦ ๐Ÿญ๐Ÿฐ. ๐˜ฟ๐™ค๐™ช๐™—๐™ก๐™ž๐™ฃ๐™œ ๐™‹๐™–๐™ฉ๐™ฉ๐™š๐™ง๐™ฃ๐™จ

Suppose a quantity doubles every 3 hours.

If ๐’• represents hours, the model is:

๐‘จ(๐’•) = ๐‘จ₀(๐Ÿฎ)แต—⁄³

Why is the exponent ๐’•⁄๐Ÿฏ?

Because one doubling occurs every 3 hours.

For example, if the starting amount is 100:

After 3 hours:

๐Ÿญ๐Ÿฌ๐Ÿฌ × ๐Ÿฎ = ๐Ÿฎ๐Ÿฌ๐Ÿฌ

After 6 hours:

๐Ÿญ๐Ÿฌ๐Ÿฌ × ๐Ÿฎ² = ๐Ÿฐ๐Ÿฌ๐Ÿฌ

After 9 hours:

๐Ÿญ๐Ÿฌ๐Ÿฌ × ๐Ÿฎ³ = ๐Ÿด๐Ÿฌ๐Ÿฌ


✦ ๐Ÿญ๐Ÿฑ. ๐™๐™ง๐™ž๐™ฅ๐™ก๐™ž๐™ฃ๐™œ ๐™‹๐™–๐™ฉ๐™ฉ๐™š๐™ง๐™ฃ๐™จ

If a quantity triples every 4 hours:

๐‘จ(๐’•) = ๐‘จ₀(๐Ÿฏ)แต—⁄⁴

If the initial value is 50:

๐‘จ(๐’•) = ๐Ÿฑ๐Ÿฌ(๐Ÿฏ)แต—⁄⁴

After 4 hours:

๐Ÿฑ๐Ÿฌ × ๐Ÿฏ = ๐Ÿญ๐Ÿฑ๐Ÿฌ

After 8 hours:

๐Ÿฑ๐Ÿฌ × ๐Ÿฏ² = ๐Ÿฐ๐Ÿฑ๐Ÿฌ


✦ ๐Ÿญ๐Ÿฒ. ๐™ƒ๐™–๐™ก๐™›-๐™‡๐™ž๐™›๐™š ๐™‹๐™–๐™ฉ๐™ฉ๐™š๐™ง๐™ฃ๐™จ

If a quantity is reduced to half every 5 years:

๐‘จ(๐’•) = ๐‘จ₀(๐Ÿญ⁄๐Ÿฎ)แต—⁄⁵

Suppose:

๐‘จ₀ = ๐Ÿญ๐Ÿฒ๐Ÿฌ

Then:

After 5 years:

๐Ÿญ๐Ÿฒ๐Ÿฌ × ๐Ÿญ⁄๐Ÿฎ = ๐Ÿด๐Ÿฌ

After 10 years:

๐Ÿญ๐Ÿฒ๐Ÿฌ × (๐Ÿญ⁄๐Ÿฎ)² = ๐Ÿฐ๐Ÿฌ

After 15 years:

๐Ÿญ๐Ÿฒ๐Ÿฌ × (๐Ÿญ⁄๐Ÿฎ)³ = ๐Ÿฎ๐Ÿฌ

The quantity keeps being multiplied by the same factor.


✦ ๐Ÿญ๐Ÿณ. ๐™’๐™๐™š๐™ฃ ๐™๐™๐™š ๐™๐™ž๐™ข๐™š ๐™๐™ฃ๐™ž๐™ฉ ๐˜พ๐™๐™–๐™ฃ๐™œ๐™š๐™จ

Be careful when the time unit in the question does not match the time unit in the model.

Suppose a quantity doubles every:

4 years

and ๐’• is measured in years.

Then:

๐‘จ(๐’•) = ๐‘จ₀(๐Ÿฎ)แต—⁄⁴

But if ๐’• represents four-year periods instead, the model could simply be:

๐‘จ(๐’•) = ๐‘จ₀(๐Ÿฎแต—)

Always ask:

“๐™’๐™๐™–๐™ฉ ๐™™๐™ค๐™š๐™จ ๐Ÿญ ๐™ช๐™ฃ๐™ž๐™ฉ ๐™ค๐™› ๐’™ ๐™ง๐™š๐™ฅ๐™ง๐™š๐™จ๐™š๐™ฃ๐™ฉ?”

That one question can prevent a major modeling error.


✦ ๐Ÿญ๐Ÿด. ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ ๐™๐™ช๐™ก๐™š๐™จ

Exponential functions become much easier when the basic exponent rules are automatic.

๐’‚⁰ = ๐Ÿญ

๐’‚แต × ๐’‚โฟ = ๐’‚แต⁺โฟ

๐’‚แต ÷ ๐’‚โฟ = ๐’‚แต⁻โฟ

(๐’‚แต)โฟ = ๐’‚แตโฟ

๐’‚⁻โฟ = ๐Ÿญ⁄๐’‚โฟ

For example:

๐Ÿฎ³ × ๐Ÿฎ⁴ = ๐Ÿฎ⁷

because:

๐Ÿฏ + ๐Ÿฐ = ๐Ÿณ


⚠️ ✦ ๐Ÿญ๐Ÿต. ๐˜ฟ๐™ค ๐™‰๐™ค๐™ฉ ๐˜ผ๐™™๐™™ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™จ ๐™’๐™๐™š๐™ฃ ๐˜ผ๐™™๐™™๐™ž๐™ฃ๐™œ

This rule:

๐’‚แต × ๐’‚โฟ = ๐’‚แต⁺โฟ

is for multiplication.

It does not mean:

๐’‚แต + ๐’‚โฟ = ๐’‚แต⁺โฟ

For example:

๐Ÿฎ² + ๐Ÿฎ³

equals:

๐Ÿฐ + ๐Ÿด = ๐Ÿญ๐Ÿฎ

It does not equal:

๐Ÿฎ⁵

Always look at the operation before choosing an exponent rule.


✦ ๐Ÿฎ๐Ÿฌ. ๐™Ž๐™ค๐™ก๐™ซ๐™ž๐™ฃ๐™œ ๐™Ž๐™ž๐™ข๐™ฅ๐™ก๐™š ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™€๐™ฆ๐™ช๐™–๐™ฉ๐™ž๐™ค๐™ฃ๐™จ

Consider:

๐Ÿฎหฃ = ๐Ÿฏ๐Ÿฎ

Rewrite 32 as a power of 2:

๐Ÿฏ๐Ÿฎ = ๐Ÿฎ⁵

Therefore:

๐Ÿฎหฃ = ๐Ÿฎ⁵

So:

๐’™ = ๐Ÿฑ

The key strategy is:

๐™๐™ง๐™ฎ ๐™ฉ๐™ค ๐™ฌ๐™ง๐™ž๐™ฉ๐™š ๐™—๐™ค๐™ฉ๐™ ๐™จ๐™ž๐™™๐™š๐™จ ๐™ฌ๐™ž๐™ฉ๐™ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™—๐™–๐™จ๐™š.


✦ ๐Ÿฎ๐Ÿญ. ๐™‚๐™ง๐™–๐™ฅ๐™๐™จ ๐™Š๐™› ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ๐™จ

For:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

the graph is curved rather than a straight line.

If:

๐’ƒ > ๐Ÿญ

the graph rises as ๐’™ increases.

If:

๐Ÿฌ < ๐’ƒ < ๐Ÿญ

the graph falls as ๐’™ increases.

The graph passes through:

(๐Ÿฌ, ๐’‚)

because:

๐’‡(๐Ÿฌ) = ๐’‚

For the basic form with no vertical shift, the graph approaches:

๐’š = ๐Ÿฌ

as the curve extends in the appropriate direction.


✦ ๐Ÿฎ๐Ÿฎ. ๐™’๐™๐™–๐™ฉ ๐˜ฟ๐™ค๐™š๐™จ ๐’Œ ๐˜ฟ๐™ค?

Consider:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ) + ๐’Œ

The +๐’Œ moves the entire graph vertically.

For example:

๐’‡(๐’™) = ๐Ÿฏ(๐Ÿฎหฃ) + ๐Ÿฑ

has horizontal asymptote:

๐’š = ๐Ÿฑ

The vertical shift changes the long-term position of the graph.


✦ ๐Ÿฎ๐Ÿฏ. ๐™๐™ž๐™ฃ๐™™๐™ž๐™ฃ๐™œ ๐™๐™๐™š ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™๐™–๐™˜๐™ฉ๐™ค๐™ง

Suppose a quantity changes from:

๐Ÿฎ๐Ÿฌ๐Ÿฌ → ๐Ÿฎ๐Ÿฏ๐Ÿฌ

The multiplier is:

๐Ÿฎ๐Ÿฏ๐Ÿฌ ÷ ๐Ÿฎ๐Ÿฌ๐Ÿฌ = ๐Ÿญ.๐Ÿญ๐Ÿฑ

Therefore the growth factor is:

๐Ÿญ.๐Ÿญ๐Ÿฑ

The percentage increase is:

๐Ÿญ.๐Ÿญ๐Ÿฑ − ๐Ÿญ = ๐Ÿฌ.๐Ÿญ๐Ÿฑ

which is:

๐Ÿญ๐Ÿฑ%

So:

×๐Ÿญ.๐Ÿญ๐Ÿฑ = ๐Ÿญ๐Ÿฑ% growth


✦ ๐Ÿฎ๐Ÿฐ. ๐™๐™ž๐™ฃ๐™™๐™ž๐™ฃ๐™œ ๐™๐™๐™š ๐˜ฟ๐™š๐™˜๐™–๐™ฎ ๐™๐™–๐™ฉ๐™š

Suppose a quantity changes from:

๐Ÿฑ๐Ÿฌ๐Ÿฌ → ๐Ÿฐ๐Ÿฎ๐Ÿฑ

The multiplier is:

๐Ÿฐ๐Ÿฎ๐Ÿฑ ÷ ๐Ÿฑ๐Ÿฌ๐Ÿฌ = ๐Ÿฌ.๐Ÿด๐Ÿฑ

The amount remaining is:

๐Ÿด๐Ÿฑ%

Therefore the decrease is:

๐Ÿญ๐Ÿฑ%

So:

×๐Ÿฌ.๐Ÿด๐Ÿฑ = ๐Ÿญ๐Ÿฑ% decay


✦ ๐Ÿฎ๐Ÿฑ. ๐™๐™๐™š ๐™Š๐™ฃ๐™š-๐™Ž๐™ฉ๐™š๐™ฅ ๐™ˆ๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ž๐™š๐™ง ๐™๐™ช๐™ก๐™š

For:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

we can write:

๐’‡(๐’™ + ๐Ÿญ) = ๐’ƒ๐’‡(๐’™)

This is powerful because it tells you exactly what happens after one additional step.

Suppose:

๐’‡(๐’™ + ๐Ÿญ) = ๐Ÿฐ๐’‡(๐’™)

Then the multiplier is:

๐’ƒ = ๐Ÿฐ

If:

๐’‡(๐Ÿฌ) = ๐Ÿฏ

then:

๐’‡(๐Ÿญ) = ๐Ÿญ๐Ÿฎ

๐’‡(๐Ÿฎ) = ๐Ÿฐ๐Ÿด

๐’‡(๐Ÿฏ) = ๐Ÿญ๐Ÿต๐Ÿฎ


✦ ๐Ÿฎ๐Ÿฒ. ๐™Ž๐˜ผ๐™ ๐™’๐™ค๐™ง๐™™ ๐™‹๐™ง๐™ค๐™—๐™ก๐™š๐™ข ๐™๐™ง๐™–๐™ฃ๐™จ๐™ก๐™–๐™ฉ๐™ž๐™ค๐™ฃ

Words such as these should immediately make you think about exponential models:

“increases by ๐Ÿฑ% each year”

×๐Ÿญ.๐Ÿฌ๐Ÿฑ

“decreases by ๐Ÿญ๐Ÿฎ% each month”

×๐Ÿฌ.๐Ÿด๐Ÿด

“doubles every ๐Ÿฏ hours”

×๐Ÿฎ every ๐Ÿฏ hours

“triples every ๐Ÿฑ days”

×๐Ÿฏ every ๐Ÿฑ days

“is reduced by half every ๐Ÿฐ years”

×๐Ÿญ⁄๐Ÿฎ every ๐Ÿฐ years

The wording changes.

The underlying mathematics remains the same.


✦ ๐Ÿฎ๐Ÿณ. ๐™’๐™ค๐™ง๐™™ ๐™‹๐™ง๐™ค๐™—๐™ก๐™š๐™ข: ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™

A town has a population of ๐Ÿญ๐Ÿฌ,๐Ÿฌ๐Ÿฌ๐Ÿฌ and grows by ๐Ÿฎ% each year.

Step ๐Ÿญ: Starting value

๐’‚ = ๐Ÿญ๐Ÿฌ๐Ÿฌ๐Ÿฌ๐Ÿฌ

Step ๐Ÿฎ: Growth factor

๐Ÿญ + ๐Ÿฌ.๐Ÿฌ๐Ÿฎ = ๐Ÿญ.๐Ÿฌ๐Ÿฎ

Step ๐Ÿฏ: Build the model

๐‘ท(๐’•) = ๐Ÿญ๐Ÿฌ๐Ÿฌ๐Ÿฌ๐Ÿฌ(๐Ÿญ.๐Ÿฌ๐Ÿฎแต—)

The equation describes the population after ๐’• years.


✦ ๐Ÿฎ๐Ÿด. ๐™’๐™ค๐™ง๐™™ ๐™‹๐™ง๐™ค๐™—๐™ก๐™š๐™ข: ๐˜ฟ๐™š๐™˜๐™–๐™ฎ

A car is worth $๐Ÿฎ๐Ÿฌ,๐Ÿฌ๐Ÿฌ๐Ÿฌ and loses ๐Ÿญ๐Ÿฑ% of its value each year.

Remaining percentage:

๐Ÿญ๐Ÿฌ๐Ÿฌ% − ๐Ÿญ๐Ÿฑ% = ๐Ÿด๐Ÿฑ%

Multiplier:

๐Ÿฌ.๐Ÿด๐Ÿฑ

Therefore:

๐‘ฝ(๐’•) = ๐Ÿฎ๐Ÿฌ๐Ÿฌ๐Ÿฌ๐Ÿฌ(๐Ÿฌ.๐Ÿด๐Ÿฑแต—)

Notice that the exponent counts the number of years.


✦ ๐Ÿฎ๐Ÿต. ๐™๐™๐™š ๐™Ž๐™–๐™ข๐™š ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š ๐˜ฟ๐™ค๐™š๐™จ ๐™‰๐™ค๐™ฉ ๐™ˆ๐™š๐™–๐™ฃ ๐™๐™๐™š ๐™Ž๐™–๐™ข๐™š ๐˜ผ๐™ข๐™ค๐™ช๐™ฃ๐™ฉ

This is a crucial concept.

Suppose a value is:

๐Ÿญ๐Ÿฌ๐Ÿฌ

and decreases by ๐Ÿญ๐Ÿฌ%.

First decrease:

๐Ÿญ๐Ÿฌ๐Ÿฌ × ๐Ÿฌ.๐Ÿต = ๐Ÿต๐Ÿฌ

Second decrease:

๐Ÿต๐Ÿฌ × ๐Ÿฌ.๐Ÿต = ๐Ÿด๐Ÿญ

Third decrease:

๐Ÿด๐Ÿญ × ๐Ÿฌ.๐Ÿต = ๐Ÿณ๐Ÿฎ.๐Ÿต

The decrease amounts are:

๐Ÿญ๐Ÿฌ

then:

๐Ÿต

then:

๐Ÿด.๐Ÿญ

The percentage remains the same.

The actual amount changes.

That is why the process is exponential.


✦ ๐Ÿฏ๐Ÿฌ. ๐™’๐™๐™ฎ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐˜พ๐™–๐™ฃ ๐™Ž๐™ช๐™ง๐™ฅ๐™–๐™จ๐™จ ๐™‡๐™ž๐™ฃ๐™š๐™–๐™ง ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™

Imagine:

Linear: add 10 each step.

Exponential: multiply by 1.10 each step.

Starting from 100:

Linear:

๐Ÿญ๐Ÿฌ๐Ÿฌ → ๐Ÿญ๐Ÿญ๐Ÿฌ → ๐Ÿญ๐Ÿฎ๐Ÿฌ → ๐Ÿญ๐Ÿฏ๐Ÿฌ → ๐Ÿญ๐Ÿฐ๐Ÿฌ

Exponential:

๐Ÿญ๐Ÿฌ๐Ÿฌ → ๐Ÿญ๐Ÿญ๐Ÿฌ → ๐Ÿญ๐Ÿฎ๐Ÿญ → ๐Ÿญ๐Ÿฏ๐Ÿฏ.๐Ÿญ → ๐Ÿญ๐Ÿฐ๐Ÿฒ.๐Ÿฐ๐Ÿญ

At first the values look similar.

But repeated multiplication can eventually produce a very large difference.


✦ ๐Ÿฏ๐Ÿญ. ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐˜พ๐™–๐™ฃ ๐˜ฝ๐™š ๐™๐™–๐™จ๐™ฉ

Consider:

๐’‡(๐’™) = ๐Ÿฎ(๐Ÿฏหฃ)

The first few values are:

๐’™ = ๐Ÿฌ → ๐Ÿฎ

๐’™ = ๐Ÿญ → ๐Ÿฒ

๐’™ = ๐Ÿฎ → ๐Ÿญ๐Ÿด

๐’™ = ๐Ÿฏ → ๐Ÿฑ๐Ÿฐ

๐’™ = ๐Ÿฐ → ๐Ÿญ๐Ÿฒ๐Ÿฎ

Every step multiplies the previous output by 3.

That repeated multiplication is the heart of exponential growth.


✦ ๐Ÿฏ๐Ÿฎ. ๐™๐™๐™š ๐™ˆ๐™ค๐™จ๐™ฉ ๐˜พ๐™ค๐™ข๐™ข๐™ค๐™ฃ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š๐™จ

❌ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š ๐Ÿญ: ๐™๐™จ๐™ž๐™ฃ๐™œ ๐Ÿฌ.๐Ÿฌ๐Ÿฑ ๐™๐™ค๐™ง ๐Ÿฑ% ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™

Correct:

๐Ÿญ.๐Ÿฌ๐Ÿฑ


❌ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š ๐Ÿฎ: ๐™๐™จ๐™ž๐™ฃ๐™œ ๐Ÿฌ.๐Ÿฎ๐Ÿฌ ๐™๐™ค๐™ง ๐Ÿฎ๐Ÿฌ% ๐˜ฟ๐™š๐™˜๐™ง๐™š๐™–๐™จ๐™š

Correct:

๐Ÿฌ.๐Ÿด๐Ÿฌ


❌ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š ๐Ÿฏ: ๐˜พ๐™๐™š๐™˜๐™ ๐™ž๐™ฃ๐™œ ๐˜ฟ๐™ž๐™›๐™›๐™š๐™ง๐™š๐™ฃ๐™˜๐™š๐™จ ๐™„๐™ฃ๐™จ๐™ฉ๐™š๐™–๐™™ ๐™Š๐™› ๐™๐™–๐™ฉ๐™ž๐™ค๐™จ

For exponential tables, divide consecutive values.


❌ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š ๐Ÿฐ: ๐™„๐™œ๐™ฃ๐™ค๐™ง๐™ž๐™ฃ๐™œ ๐™๐™ž๐™ข๐™š ๐™๐™ฃ๐™ž๐™ฉ๐™จ

“Doubles every 5 years” does not mean it doubles every year.


❌ ๐™ˆ๐™ž๐™จ๐™ฉ๐™–๐™ ๐™š ๐Ÿฑ: ๐™ˆ๐™ž๐™ญ๐™ž๐™ฃ๐™œ ๐™๐™ฅ ๐™Ž๐™ฉ๐™–๐™ง๐™ฉ๐™ž๐™ฃ๐™œ ๐™‘๐™–๐™ก๐™ช๐™š ๐˜ผ๐™ฃ๐™™ ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™๐™–๐™˜๐™ฉ๐™ค๐™ง

In:

๐’‡(๐’™) = ๐Ÿฑ๐Ÿฌ(๐Ÿญ.๐Ÿฌ๐Ÿฐหฃ)

50 is the starting value.

1.04 is the growth factor.


✦ ๐Ÿฏ๐Ÿฏ. ๐™๐™–๐™จ๐™ฉ ๐™Ž๐™–๐™ฉ ๐™Ž๐™ฉ๐™ง๐™–๐™ฉ๐™š๐™œ๐™ฎ

When you see an exponential question, stop before calculating.

Ask these questions:

① ๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐™ฉ๐™๐™š ๐™จ๐™ฉ๐™–๐™ง๐™ฉ๐™ž๐™ฃ๐™œ ๐™ซ๐™–๐™ก๐™ช๐™š?

② ๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐™ฉ๐™๐™š ๐™ข๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ž๐™š๐™ง?

③ ๐™„๐™จ ๐™ž๐™ฉ ๐™œ๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™ค๐™ง ๐™™๐™š๐™˜๐™–๐™ฎ?

④ ๐™’๐™๐™–๐™ฉ ๐™™๐™ค๐™š๐™จ ๐’™ ๐™ง๐™š๐™ฅ๐™ง๐™š๐™จ๐™š๐™ฃ๐™ฉ?

⑤ ๐™ƒ๐™ค๐™ฌ ๐™ข๐™–๐™ฃ๐™ฎ ๐™ฉ๐™ž๐™ข๐™š๐™จ ๐™ž๐™จ ๐™ฉ๐™๐™š ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง ๐™–๐™ฅ๐™ฅ๐™ก๐™ž๐™š๐™™?

If you answer those five questions, many apparently difficult problems become much simpler.


✦ ๐Ÿฏ๐Ÿฐ. ๐™Œ๐™ช๐™ž๐™˜๐™  ๐™Ž๐˜ผ๐™ ๐™‹๐™ง๐™–๐™˜๐™ฉ๐™ž๐™˜๐™š

๐™Œ๐™ช๐™š๐™จ๐™ฉ๐™ž๐™ค๐™ฃ ๐Ÿญ

Which equation represents exponential growth?

๐˜ผ) ๐’š = ๐Ÿฏ๐’™ + ๐Ÿฎ

๐˜ฝ) ๐’š = ๐Ÿฏ๐’™² + ๐Ÿฎ

๐˜พ) ๐’š = ๐Ÿฏ(๐Ÿญ.๐Ÿฑหฃ)

๐˜ฟ) ๐’š = ๐Ÿฏ⁄๐’™

๐˜ผ๐™ฃ๐™จ๐™ฌ๐™š๐™ง: ๐˜พ

The variable appears in the exponent and the base is greater than 1.


๐™Œ๐™ช๐™š๐™จ๐™ฉ๐™ž๐™ค๐™ฃ ๐Ÿฎ

A quantity increases by 12% each year.

What is the growth factor?

๐˜ผ๐™ฃ๐™จ๐™ฌ๐™š๐™ง: ๐Ÿญ.๐Ÿญ๐Ÿฎ

Because:

๐Ÿญ + ๐Ÿฌ.๐Ÿญ๐Ÿฎ = ๐Ÿญ.๐Ÿญ๐Ÿฎ


๐™Œ๐™ช๐™š๐™จ๐™ฉ๐™ž๐™ค๐™ฃ ๐Ÿฏ

A quantity decreases by 35% each month.

What multiplier should be used?

๐˜ผ๐™ฃ๐™จ๐™ฌ๐™š๐™ง: ๐Ÿฌ.๐Ÿฒ๐Ÿฑ

Because:

๐Ÿญ − ๐Ÿฌ.๐Ÿฏ๐Ÿฑ = ๐Ÿฌ.๐Ÿฒ๐Ÿฑ


๐™Œ๐™ช๐™š๐™จ๐™ฉ๐™ž๐™ค๐™ฃ ๐Ÿฐ

The values in a table are:

๐Ÿฐ, ๐Ÿญ๐Ÿฎ, ๐Ÿฏ๐Ÿฒ, ๐Ÿญ๐Ÿฌ๐Ÿด

What is the common ratio?

๐˜ผ๐™ฃ๐™จ๐™ฌ๐™š๐™ง: ๐Ÿฏ

because:

๐Ÿญ๐Ÿฎ ÷ ๐Ÿฐ = ๐Ÿฏ

๐Ÿฏ๐Ÿฒ ÷ ๐Ÿญ๐Ÿฎ = ๐Ÿฏ

๐Ÿญ๐Ÿฌ๐Ÿด ÷ ๐Ÿฏ๐Ÿฒ = ๐Ÿฏ


๐™Œ๐™ช๐™š๐™จ๐™ฉ๐™ž๐™ค๐™ฃ ๐Ÿฑ

A quantity starts at 80 and doubles every 4 hours.

What is its value after 12 hours?

There are:

๐Ÿญ๐Ÿฎ ÷ ๐Ÿฐ = ๐Ÿฏ

doubling periods.

Therefore:

๐Ÿด๐Ÿฌ × ๐Ÿฎ³

= ๐Ÿด๐Ÿฌ × ๐Ÿด

= ๐Ÿฒ๐Ÿฐ๐Ÿฌ


✦ ๐Ÿฏ๐Ÿฑ. ๐™๐™๐™š ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ ๐™ˆ๐™š๐™ข๐™ค๐™ง๐™ฎ ๐˜พ๐™–๐™ง๐™™

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)

๐’‚ → starting value

๐’ƒ → repeated multiplier

๐’™ → number of steps

๐’ƒ > ๐Ÿญ → growth

๐Ÿฌ < ๐’ƒ < ๐Ÿญ → decay

๐’‡(๐Ÿฌ) = ๐’‚

๐’“% increase → ×(๐Ÿญ + ๐’“)

๐’“% decrease → ×(๐Ÿญ − ๐’“)

constant difference → linear

constant ratio → exponential

double → ×๐Ÿฎ

triple → ×๐Ÿฏ

half → ×๐Ÿญ⁄๐Ÿฎ

๐’‡(๐’™ + ๐Ÿญ) = ๐’ƒ๐’‡(๐’™)


✦ ๐Ÿฏ๐Ÿฒ. ๐™Ž๐˜ผ๐™ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ๐™จ ๐™๐˜ผ๐™Œ

๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐™–๐™ฃ ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™›๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ?

A function in which the variable appears in the exponent, commonly written:

๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)


๐™ƒ๐™ค๐™ฌ ๐™™๐™ค ๐™„ ๐™ž๐™™๐™š๐™ฃ๐™ฉ๐™ž๐™›๐™ฎ ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™œ๐™ง๐™ค๐™ฌ๐™ฉ๐™?

Look at the base.

If:

๐’ƒ > ๐Ÿญ

the function grows.


๐™ƒ๐™ค๐™ฌ ๐™™๐™ค ๐™„ ๐™ž๐™™๐™š๐™ฃ๐™ฉ๐™ž๐™›๐™ฎ ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™™๐™š๐™˜๐™–๐™ฎ?

If:

๐Ÿฌ < ๐’ƒ < ๐Ÿญ

the function decays.


๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐Ÿฎ๐Ÿฌ% ๐™œ๐™ง๐™ค๐™ฌ๐™ฉ๐™ ๐™–๐™จ ๐™– ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง?

๐Ÿญ.๐Ÿฎ


๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐Ÿฎ๐Ÿฌ% ๐™™๐™š๐™˜๐™–๐™ฎ ๐™–๐™จ ๐™– ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง?

๐Ÿฌ.๐Ÿด


๐™ƒ๐™ค๐™ฌ ๐™™๐™ค ๐™„ ๐™›๐™ž๐™ฃ๐™™ ๐™–๐™ฃ ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™ฅ๐™–๐™ฉ๐™ฉ๐™š๐™ง๐™ฃ ๐™ž๐™ฃ ๐™– ๐™ฉ๐™–๐™—๐™ก๐™š?

Divide consecutive output values.

If the ratios remain the same, the pattern is exponential.


๐™’๐™๐™–๐™ฉ ๐™ž๐™จ ๐™ฉ๐™๐™š ๐™™๐™ž๐™›๐™›๐™š๐™ง๐™š๐™ฃ๐™˜๐™š ๐™—๐™š๐™ฉ๐™ฌ๐™š๐™š๐™ฃ ๐™š๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™–๐™ฃ๐™™ ๐™ก๐™ž๐™ฃ๐™š๐™–๐™ง ๐™œ๐™ง๐™ค๐™ฌ๐™ฉ๐™?

Linear growth repeatedly adds the same amount.

Exponential growth repeatedly multiplies by the same factor.


๐™’๐™๐™–๐™ฉ ๐™™๐™ค๐™š๐™จ ๐’‚ ๐™ข๐™š๐™–๐™ฃ ๐™ž๐™ฃ ๐’‡(๐’™) = ๐’‚(๐’ƒหฃ)?

It is the starting value because:

๐’‡(๐Ÿฌ) = ๐’‚


๐™’๐™๐™–๐™ฉ ๐™™๐™ค๐™š๐™จ ๐’ƒ ๐™ข๐™š๐™–๐™ฃ?

It is the multiplier applied whenever ๐’™ increases by one unit.


✦ ๐™๐™ž๐™ฃ๐™–๐™ก ๐™๐™–๐™ ๐™š๐™–๐™ฌ๐™–๐™ฎ

The easiest way to understand exponential functions is not to memorize dozens of separate examples.

Understand the pattern:

๐™‡๐™ž๐™ฃ๐™š๐™–๐™ง → ๐™–๐™™๐™™ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™–๐™ข๐™ค๐™ช๐™ฃ๐™ฉ.

๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก → ๐™ข๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ฎ ๐™—๐™ฎ ๐™ฉ๐™๐™š ๐™จ๐™–๐™ข๐™š ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง.

When you see an exponential question, find:

๐™ฉ๐™๐™š ๐™จ๐™ฉ๐™–๐™ง๐™ฉ๐™ž๐™ฃ๐™œ ๐™ซ๐™–๐™ก๐™ช๐™š

๐™ฉ๐™๐™š ๐™ข๐™ช๐™ก๐™ฉ๐™ž๐™ฅ๐™ก๐™ž๐™š๐™ง

๐™ฉ๐™๐™š ๐™ฉ๐™ž๐™ข๐™š ๐™ž๐™ฃ๐™ฉ๐™š๐™ง๐™ซ๐™–๐™ก

๐™–๐™ฃ๐™™ ๐™ฉ๐™๐™š ๐™ฃ๐™ช๐™ข๐™—๐™š๐™ง ๐™ค๐™› ๐™ง๐™š๐™ฅ๐™š๐™–๐™ฉ๐™š๐™™ ๐™˜๐™๐™–๐™ฃ๐™œ๐™š๐™จ.

Once those four pieces are clear, the equation usually becomes much easier to see.

๐™Ž๐™š๐™š ๐™ฉ๐™๐™š ๐™›๐™–๐™˜๐™ฉ๐™ค๐™ง. ๐™๐™š๐™˜๐™ค๐™œ๐™ฃ๐™ž๐™ฏ๐™š ๐™ฉ๐™๐™š ๐™ฅ๐™–๐™ฉ๐™ฉ๐™š๐™ง๐™ฃ. ๐™๐™๐™š๐™ฃ ๐™จ๐™ค๐™ก๐™ซ๐™š.

other pages to explore

SAT MATH FORMULA  SHEET FOR QUICK REFERENCE

SAT MATH FORMULA CHEAT SHEET 


ALGEBRA

SO;VING LINEAR EQUATIONS [PART 1]

LINEAR EQUATIONS [PART 1]


SOLVING LINEAR EQUATIONS [PART II]

LINEAR EQUATIONS [PART II]


SYSTEM OF EQUATIONS  [ PART  I ]


SAT Systems of Equations: Part 2 — Hard Questions, Word Problems, Graphs, Parameters & SAT Tricks

SAT system of equations 2


 PARAMETER QUESTIONS IN SYSTEMS OF LINEAR EQUATIONS


QUADRATIC EQUATIONS [PART I] 

QUADRATIC EQUATIONS [PART I] 


QUADRATIC EQUATIONS [PART II] 

QUADRATIC EQUATIONS [PART II] 


Linear Inequality

linear inequalities


PERCENTAGES


PERCENTAGES [introduction]

PERCENTAGES [part 1]


PERCENTAGE INCREASE AND DECREASE

PERCENTAGES [part 2]


SUCCESSIVE PERCENTAGES DISCOUNT PROFIT LOSS

PERCENTAGES [part3]

Percentages in Data Analysis, Graphs, Tables, Probability,SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions and Advanced Word Problems


SAT Reverse Percentage Problems




GEOMETRY

SAT Geometry Notes:  Study Guide with Formulas, Tricks, Practice Questions





SAT Word Problems Made Easy: Tricks to Solve Digital SAT Math Problems Faster



The Big Word-Problem Families:  SAT Math Guide


How SAT Word Problems Hide the Equation: Skill That Makes Difficult Questions Easier



SAT Word Problems Distance, Speed, Work, Age, Mixtures & Profit — A Smarter Way to Decode the Question



SAT Exponential Functions: Growth, Decay, Formulas & Easy Tricks


SAT Function Transformations: A Simple Way to Master Shifts, Reflections, Stretches and Compressions





SAT Probability Made Easy: Conditional Probability, Tables & Tricky Questions




SAT Conditional Probability

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SAT Exponential Functions: Growth, Decay, Formulas & Easy Tricks

๐™Ž๐˜ผ๐™ ๐™€๐™ญ๐™ฅ๐™ค๐™ฃ๐™š๐™ฃ๐™ฉ๐™ž๐™–๐™ก ๐™๐™ช๐™ฃ๐™˜๐™ฉ๐™ž๐™ค๐™ฃ๐™จ ๐™‚๐™ช๐™ž๐™™๐™š ๐™‚๐™ง๐™ค๐™ฌ๐™ฉ๐™, ๐˜ฟ๐™š๐™˜๐™–๐™ฎ, ๐™‹๐™š๐™ง๐™˜๐™š๐™ฃ๐™ฉ๐™–๐™œ๐™š๐™จ, ๐™‚๐™ง๐™–๐™ฅ๐™๐™จ, ๐™๐™–๐™—๐™ก๐™š๐™จ ๐™–...