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.



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Comments 4 comments

tlpoague profile image

tlpoague 6 years ago from USA

I am horrible a math problem solving, but found this way at looking at logarithms interesting. Thanks for sharing.


cristina327 profile image

cristina327 6 years ago from Manila Author

Hi tipoague thank you for dropping by. I am glad you find this interesting. Blessings to you and regards.


KellyAkhil profile image

KellyAkhil 6 years ago from India,Hyderabad

excellent..... you are really great.you are posting Quite interesting hubs regarding Mathematics its nice.

Can you give some Short Techniques for calculating Time&work,Heights&distances,Probability,Number work. Please publish these hubs. Really i like your hubs. Thanks for giving this informative hubs.


andrew 4 years ago

hey in that first problem with the coffee, in your origional equation at the top, you have -30t instead of -.30t. this caused me a few minutes of confusion, but the rest of the problem was solved correctly.

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