Application of Logarithm

The following famous saying of mathematical Laplace gives crux of application of logarithm in mathematics. "The invention of logarithms shortens the calculations extending over months to just a few days and thereby as it were doubles the life span of calculator."
Now we will see how it shortens the calculations.

Examples on application of logarithm

1) If \$1750 is invested at 9% interest per year for 10 years. Find a) the interest compounded annually. b) the interest compounded half yearly and , c) the difference between (a) and (b).
Solution : Using the formula,
A = P(1 + r)
t
∴ A = 1750 (1 + 0.09)
10
log A = log(1750) + 10 log(1.09)
= 3.2430 + 10 x 0.03740
= 3.2430 + 0.3740
log A = 3.6170
∴ A = 4139.996 = \$ 4140
(a) Interest = 4140 - 1750 = \$2390
(b) the interest compounded half yearly
A = P(1 + r/2)
2t
∴ A = 1750 (1 + 0.09/2)
20
∴ A = 1750 (1 + 0.045)
20
log A = log(1750) + 10 log(1.045)
= 3.2430 + 20 x 0.01911
= 3.2430 + 0.382325
log A = 3.6250
∴ A = \$4216.96 = \$ 4217
(a) Interest = 4217 - 1750 = \$2467
(c) Difference between (a) and (b) = \$2467 - \$2390 = \$77
∴ The interest compounded half yearly is \$77 more that the interest compounded annually.

2) The initial bacterium count in a culture is 200. A biologist later makes a sample count of bacteria in the culture and finds that the relative rate of growth is 30% per hour.
(a) Find a function that models the number of bacteria after t hours.
(b) What is the estimated count after 1 hour?
(c) What is the estimated count after 8 hours?
Solution : We will use n 0 as initial population, n is the population after time t in hours.Rate = r (in percent)
Formula : n(t) = n0 ert
(a) n(t) = 200 e 0.3t
Since here we have used a exponential growth so we will use natural logarithm.
(b) The estimated count after 1 hour
n(1) = ln (200) + 0.3 ln(e) ----(For t = 1 hour )
n(1) = 5.29831 + 0.3----(since ln(e) = 1)
n(1) = 5.59831
n(1) = 269.96 = 270
(c) The estimated count after 8 hours
n(8) = ln (200) + 2.4 ln(e) ----(since t = 8 hour )
n(8) = 5.29831 + 2.4(since ln(e) = 1)
n(8) = 7.69831
n(8) = 2204.619 = 2204.62