Exponential and Logarithmic Functions

[sorayasem]

I have literally been trying to do this for the past 5 days and can't understand it, I'm lost. Could I get help ?

Election returns are broadcast in a town of 1 million people, and the number of people who have heard the news within t hours is 1,000,000(1-e^0.3t). How long will it take for 800,000 people to hear the news? Round your answer to the nearest number of hours.

Thank you

 

[JawadLeNain]

You should use the elasticity formula with the exponential law.

 

[lev888]

We are given an expression with the variable t, the result of which is the number of people corresponding to t hours. If we know the time t we can calculate the number of people. But what do we do if we know the number of people and we need to calculate the time?

 

[lookagain]

Election returns are broadcast in a town of 1 million people, and the number of people who have heard the news within t hours is 1,000,000(1-e^0.3t). How long will it take for 800,000 people to hear the news? Round your answer to the nearest number of hours.

It's missing grouping symbols:

1,000,000[1 - e^(0.3t)]

Set it equal to the target population.

1,000,000[1 - e^(0.3t)] = 800,000

1 - e^(0.3t) = 0.8

-e^(0.3t) = -0.2

e^(0.3t) = 0.2


Can you continue?

 

[HallsofIvy]

The problem asks you to solve the equation \(\displaystyle 1000000(1- e^{0.3t})= 8000000\).

An obvious first step is to divide both sides by 1000000 to get \(\displaystyle 1- e^{0.3t}= 8\).

Subtract 1 from both sides: \(\displaystyle -e^{0.3t}= 7\).

Multiply both sides by -1: \(\displaystyle e^{0.3t}= -7\)

No wonder this caused difficulties! But it shouldn't have taken five days to realize that e to any power is never negative! There is no such "t". Are you sure you have copied the problem correctly? That "\(\displaystyle 1- e^{0.3t}\)" means that the number of people who have heard this is decreasing as t increases.

 

[sorayasem]

The problem asks you to solve the equation \(\displaystyle 1000000(1- e^{0.3t})= 8000000\).

An obvious first step is to divide both sides by 1000000 to get \(\displaystyle 1- e^{0.3t}= 8\).

Subtract 1 from both sides: \(\displaystyle -e^{0.3t}= 7\).

Multiply both sides by -1: \(\displaystyle e^{0.3t}= -7\)

No wonder this caused difficulties! But it shouldn't have taken five days to realize that e to any power is never negative! There is no such "t". Are you sure you have copied the problem correctly? That "\(\displaystyle 1- e^{0.3t}\)" means that the number of people who have heard this is decreasing as t increases.

Hello,
thank you!! yes I wrote the problem correctly. Are you sure it's 8 though and not 0.8? I tried doing it on my calculator and it should be 800,000 not 8 million at the beginning

 

[Steven G]

Yes, Prof Halls had one too many zeros.

 

[sorayasem]

It's missing grouping symbols:

1,000,000[1 - e^(0.3t)]

Set it equal to the target population.

1,000,000[1 - e^(0.3t)] = 800,000

1 - e^(0.3t) = 0.8

-e^(0.3t) = -0.2

e^(0.3t) = 0.2


Can you continue?

Do I have to subtract 1 to 0.8 or do I have to add and find the following ? :
1 000 000 (1-e^0.3t)
1 000 000 (1-e^0.3t)= 800 000
1-e^(0.3t) = 800 000/1 000 000
1-e^(0.3t) = 0.8
e^-0.3t = 1.8
ln e^-0.3t = ln e^4.8
-0.3t = 1.8
t = 1.8/-0.3
= 1.5
around 2 hours

 

[Dr.Peterson]

Do I have to subtract 1 to 0.8 or do I have to add and find the following ? :
1 000 000 (1-e^0.3t)
1 000 000 (1-e^0.3t)= 800 000
1-e^(0.3t) = 800 000/1 000 000
1-e^(0.3t) = 0.8
e^-0.3t = 1.8
ln e^-0.3t = ln e^4.8
-0.3t = 1.8
t = 1.8/-0.3
= 1.5
around 2 hours

Think about what you did there, step by step. When subtraction or negatives are involved, it is easy to make sign errors, so you need to take it slowly. I'd do this, writing at least the steps in bold:

1-e^(0.3t) = 0.8​

1-e^(0.3t)+e^(0.3t) = 0.8+e^(0.3t)​

1 = 0.8+e^(0.3t)​

1-0.8 = 0.8+e^(0.3t)-0.8​

0.2 = e^(0.3t)​

That isn't what you got, is it?

 

[Steven G]

Do I have to subtract 1 to 0.8 or do I have to add and find the following ? :
1 000 000 (1-e^0.3t)
1 000 000 (1-e^0.3t)= 800 000
1-e^(0.3t) = 800 000/1 000 000
1-e^(0.3t) = 0.8
e^-0.3t = 1.8
ln e^-0.3t = ln e^1.8
-0.3t = 1.8
t = 1.8/-0.3
= 1.5
around 2 hours

-0.3t = 1.8
t = 1.8/-0.3 = 1.5 This is not correct. 1st of all 1.8 is a positive number and -0.3 is a negative number. The division of positive and negative is negative. Well 1.5 is NOT negative. 1.8/0.3 will be the same answer as 18/3 with the decimal place possible in a different place. That is 1.8/0.3 is .... .006 or .06 or .6 or 6 or 60 or 600.... . How did you get 1.5??

You should know what the missing number is in this equation. 1 - __ = 0.8. Think of this scenario. You have $1, spend some of it and now have $0.8=$0.80. You should know that the unknown number is 0.2. That is e^(0.3t) MUST be equal to that special number that you subtract from 1 to get 0.8, namely 0.2. So e^(0.3t)= .2

e^-0.3t = 1.8
ln e^-0.3t = ln e^1.8 You did not take ln of both sides. To take ln of both sides you simply write ln in front of the expression on each side. That is if you have e^-0.3t = 1.8 and want to take ln of both sides then you get ln(e^-0.3t) = ln(1.8)

 

[lookagain]

Do I have to subtract 1 to 0.8 or do I have to add and find the following ? :
1 000 000 (1-e^0.3t)
1 000 000 (1-e^0.3t)= 800 000
1-e^(0.3t) = 800 000/1 000 000
1-e^(0.3t) = 0.8
e^-0.3t = 1.8
ln e^-0.3t = ln e^4.8
-0.3t = 1.8
t = 1.8/-0.3
= 1.5
around 2 hours

sorayasem, you still wrote steps without the required grouping symbols as I showed, not to mention
your other errors. And note Dr. Peterson's version of corrections in his post to you.

Here are some additional/extra steps of mine:

1,000,000[1 - e^(0.3t)]

Set it equal to the target population.

1,000,000[1 - e^(0.3t)] = 800,000

1 - e^(0.3t) = 0.8

1 - e^(0.3t) - 1 = 0.8 - 1

-e^(0.3t) = -0.2

(-1)[-e^(0.3t)] = (-1)(-0.2)

e^(0.3t) = 0.2

ln[e^(0.3t)] = ln(0.2)

0.3t = ln(0.2)

\(\displaystyle \dfrac{0.3t}{0.3} \ = \ \dfrac{ln(0.2)}{0.3}\)

\(\displaystyle t \ = \ \dfrac{ln(0.2)}{0.3}\)


Now, enter that into a calculator/computer.