Applications of exponential and logarithmic functions

Exponential and logarithmic functions analyze real-life problems like compound interest, depreciation, population growth/decay, radioactive decay, and cooling.

Compound interest

The formula for compound interest is:

$A_n = P\left(1 + \frac{r}{n}\right)^{nt}$

Where:

$P$ = principal

$t$ = number of years

$r$ = interest rate (as a decimal)

$n$ = number of times interest is compounded per year

$A_n$ = accumulated amount

Example 1: Doubling the principal

How many years for Mary to double her principal at 10% interest compounded annually?

Solution:

$P = P$

$A = 2P$

$r = 0.10$

$n = 1$

$2P = P\left(1 + \frac{0.10}{1}\right)^{1 \cdot t}$

$2 = (1.1)^t$

$\log 2 = t \log 1.1$

$t = \frac{\log 2}{\log 1.1}$

$t \approx 7.273$ years (7 years and ~4 months)

Example 2: Calculating accumulated amount and interest

Kaimoto deposited Tsh 5250 at 8% compounded semi-annually for 3 years. Find the accumulated amount and compound interest.

Solution:

$P = 5250$

$r = 0.08$

$n = 2$

$t = 3$

$A = 5250\left(1 + \frac{0.08}{2}\right)^{2 \cdot 3}$

$A = 5250(1.04)^6$

$A \approx$ Tsh 6642.90

$I = A - P$

$I = 6642.90 - 5250$

$I \approx$ Tsh 1392.90

Depreciation

The formula for depreciation is:

$A = P(1 - R)^t$

Where:

$P$ = initial value

$R$ = depreciation rate (as a decimal)

$t$ = number of years

$A$ = value after $t$ years

Example 3: Calculating depreciated price

A phone bought for Tsh 400,000 depreciates at 10% and 25% for the first and second years, respectively. Calculate the new price.

Solution:

$P = 400{,}000$

$R_1 = 0.10$

$R_2 = 0.25$

After 1 year: $A = 400{,}000(1 - 0.10) =$ Tsh 360,000

After 2 years: $A = 360{,}000(1 - 0.25) =$ Tsh 270,000