Solving Word Problems Involving Logarithms
Solving Word Problems Involving Logarithms
The use of logarithms has found many applications in various practical transactions of everyday life.. The use of logarithms has been very helpful in dealing with some mathematical problems encountered from time to time. In this hub, I present several word problems which involve the use of logarithm with their complete solution.
Problem Number One : Foood Source
The coding model for a coffee served in a mug is TF = TR + (TO - TR ) e^-30t.
Given that the original temperature of the coffee is 155 degrees farenheit and the room temperature is 75 degrees farenheit, determine after how many minutes the coffee will be 110 degrees farenheit ?
Given :
TF (Final Temperature) = 110 degrees farenheit
TO (Original Temperature) = 155 degrees farenheit
TR (Room Temperature) = 75 degrees farenheit
e =2.72
Formula :
TF = TR + (TO - TR) e^-.30t
Substitute the given in the formula:
110 = 75 + (155 - 75) e^-.30t
110 – 75 = ( 80) e ^-.30t
35/80 = e^-.30t
0.4375 = e^-.30t
Log 0.4375 = log e^-.30t
-0.359 = -.30t log (2.72))
-0.359 = -.30t (0.4346)
-0.359/0.4346 = -.30t
-0.826/-0.30 = t
t = 2.75 minutes
The coffee will be 110 degress farenheit after 2.75 minutes.
Problem Number Two : Sales
The model for predicting the sales S of a new brand of sweat shirt is :
S = 50,000 - 50,000 e ^-rt, where r is the rate of growth of sales. Determine the growth rate of sales to the nearest tenth of a percent if 4000 sweat shirts were sold in the fast two years.
Given :
e = 2.72
S = 4000
t = 2 years
r = ?
Formula :
S = 50,000 – 50,000 e^-rt
Substitute the given in the formula
4000 = 50,000 - 50,000 e^-2r
4000- 50,000 = -50,000e^-2r
-46,000/50,000 = - 50,000e^-2r/50,000
0.92 = e^-2r
Log 0.92 = log e^-2r
-0.0362 = -2r log log 2.72
-0.0362 = -2r 0.4346
-0.0362/0.4346 = -2r
-0.0833/-2 = r
r = 0..04 or 4 percent
Growth rate of sales is at 4 percent.
Problem Three : Finance
Solve the equation 1.05^n = 1.08^7, which shows how many years it would take for money invested at 5 percent compounded anually to equal the value of money invested at 8 percent compounded annually for seven years.
1.05^n = 1.08^7
Log 1.05^n = log 1.08^7
n log 1.05 = 7 log 1.08
n log 1.05/log 1.05 = 7 log 1.08/log 1.05
n = 7(0.0334)/0.021)
n = 11.13
It will take 11 years.
