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

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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

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

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

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

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

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

$\\ \rm\Rrightarrow t=\dfrac{log2}{log1.035}$

$\\ \rm\Rrightarrow t=20.15$

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

• So t=21years

#2

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

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

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

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

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

$\\ \rm\Rrightarrow t=\dfrac{log2.21}{log1.031}$

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

#3

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

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

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

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

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

$\\ \rm\Rrightarrow t=\dfrac{log7.5}{log1.25}$

$\\ \rm\Rrightarrow t=9.03$

Same like first case

• t=10