Question

NO LINKS!!! Exponential Growth and Decay Part 1​

Answer 1

Problem 1

a = 2500 = starting amount

b = 1 + r = 1 + 0.035 = 1.035 = growth factor

The equation is
`y = 2500*1.035^x` as it is in the form
`y = a*b^x`

Plug in y = 5000 and solve for x.

`y = 2500*1.035^x\\\\5000 = 2500*1.035^x\\\\1.035^x = 5000/2500\\\\1.035^x = 2\\\\\log(1.035^x) = \log(2)\\\\x\log(1.035) = \log(2)\\\\x = \log(2)/\log(1.035)\\\\x \approx 20.148792\\\\x \approx 21\\\\`

I rounded up to get over the hurdle. This is because plugging x = 20 will lead to y being smaller than $5000, so we must use x = 21.

Answers:

• The function is
  `y = 2500*1.035^x`
• It takes about 21 years to reach $5000

=========================================================

Problem 2

a = 4.22 = starting amount

b = 1 + r = 1 + 0.031 = 1.031 = growth factor

The equation goes from
`y = a*b^x` to
`y = 4.22*1.031^x`

Plug in y = 9.33 and solve for x.

`y = 4.22*1.031^x\\\\9.33 = 4.22*1.031^x\\\\1.031^x = 9.33/4.22\\\\1.031^x \approx 2.21090047393365\\\\\log(1.031^x) \approx \log(2.21090047393365)\\\\x\log(1.031) \approx \log(2.21090047393365)\\\\x \approx \log(2.21090047393365)/\log(1.031)\\\\x \approx 25.9882262216245\\\\x \approx 26\\\\`

Answers:

• The function is
  `y = 4.22*1.031^x`
• It takes about 26 years for the ticket to reach the price of $9.33

=========================================================

Problem 3

a = 400 = initial value

b = 1 + r = 1 + 0.25 = 1.25 = growth factor

The template
`y = a*b^x` updates to
`y = 400*1.25^x`

Plug in y = 3000 and isolate x.

`y = 400*1.25^x\\\\3000 = 400*1.25^x\\\\1.25^x = 3000/400\\\\1.25^x = 7.5\\\\\log(1.25^x) = \log(7.5)\\\\x\log(1.25) = \log(7.5)\\\\x = \log(7.5)/\log(1.25)\\\\x \approx 9.029627\\\\x \approx 10\\\\`

Like with the first problem, I rounded up to the nearest whole number. If you tried out x = 9, then y = 2980 approximately which is short of the goal of 3000. Trying x = 10 leads to y = 3725 approximately, which is now over the goal we're after.

Answers:

• The function is
  `y = 400*1.25^x`
• It takes about 10 days for 3000 people to get infected.

Answer 2

`\\ \rm\Rrightarrow P=P_o(1+r)^t`

`\\ \rm\Rrightarrow 5000=2500(1+0.035)^t`

`\\ \rm\Rrightarrow 2=1.035^t`

`\\ \rm\Rrightarrow log2=log1.035^t`

`\\ \rm\Rrightarrow log2=tlog1.035`

`\\ \rm\Rrightarrow t=(log2)/(log1.035)`

`\\ \rm\Rrightarrow t=20.15`

We can't take 20 as it can cause depict in price .

• So t=21years

#2

`\\ \rm\Rrightarrow 9.33=4.22(1+0.031)^t`

`\\ \rm\Rrightarrow 9.33=4.22(1.031)^t`

`\\ \rm\Rrightarrow 1.031^t=2.21`

`\\ \rm\Rrightarrow log1.031^t=log2.21`

`\\ \rm\Rrightarrow tlog1.031=log2.21`

`\\ \rm\Rrightarrow t=(log2.21)/(log1.031)`

`\\ \rm\Rrightarrow t=25.97\approx 26`

#3

`\\ \rm\Rrightarrow 3000=400(1+0.25)^t`

`\\ \rm\Rrightarrow 30/4=1.25^t`

`\\ \rm\Rrightarrow 7.5=1.25^t`

`\\ \rm\Rrightarrow log7.5=log1.25^t`

`\\ \rm\Rrightarrow log7.5=tlog1.25`

`\\ \rm\Rrightarrow t=(log7.5)/(log1.25)`

`\\ \rm\Rrightarrow t=9.03`

Same like first case

• t=10