Now we will see how it shortens the calculations.

A = P(1 + r)

∴ A = 1750 (1 + 0.09)

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)

∴ A = 1750 (1 + 0.09/2)

∴ A = 1750 (1 + 0.045)

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?

Since here we have used a exponential growth so we will use natural logarithm.

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

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

Application of logarithm to Home page

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