# Exponential Decay (Part 2)

How to solve exponential decay problems using logarithmics: formulas, examples, and their solutions.

## Formula

Recall the exponential growth formula:*A* = *A*_{0}(1 - *r*)^{t}

Exponential decay (part 1)

By using the logarithm,

now you can find the value of *t*.

## Example 1

Write the given values to use.

Put the given values into the formula.

100⋅(1 - 0.14)^{t} = 10

Log both sides. (base: 0.86)

Logarithmic form

To use the given condition,

change the base to 10.

Change the base formula

log (1/10) = -1

(See the right formula: log_{b} (1/*b*) = -1)

log 0.86 = -0.066

(given condition)

*t* = 15.xx weeks

But the question says 'how many weeks'.

So *t* is a natural number.

So round 15.xx up to the nearest ones:*t* = 16 weeks.

This means after 16 weeks,

the weight of the eraser becomes less than 10g.

## Formula: Continuous Decay

Recall the continuous decay formula:*A* = *A*_{0}*e*^{-rt}

Exponential decay (part 1)

By using the logarithm,

you can also find *t* from this formula.

## Example 2

Write the given values to use.

The half-life is the amount of time that takes

to change from *A*_{0} to (1/2)*A*_{0}.

So *A* = (1/2)*A*_{0}.

Put the given values into the formula.*A*_{0}⋅*e*^{-0.05⋅t} = (1/2)*A*_{0}

Log both sides. (base: *e*)

Natural logarithms

ln (1/2) = -ln 2

(See the right formula: log_{b} (1/*c*) = -log_{b} *c*)

Logarithms of powers

ln 2 = 0.69

(given condition)

This means

the amount of quantity (weight, concentration, etc.)

will be half in every 69/5 minutes (= 13.8 minutes).