Exponentials - Word Problems

An investment of $$500$ is invested in an account that pays $6.4$% compound annually. How long will it take for the original amount to triple?

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First, we can identify what each variable and value represents:

$5000$ represents the intial (or principal) amount
$1.064$ represents the Growth Factor
$t$ represents the period in years where the investment increases

Using all of the found values, we can write the equation as such:

$\text{CI} = 500(1.064)^t$

We can determine then the original amount will triple by first setting the function equal to 1500, triple the principal amount. We can then divide both sides by $500$:

$1500 = 500(1.064)^t$

$3 = (1.064)^t$

Next, we can take the common log of both sides and use the Logarithm Power Rule to bring the exponent down. Then, we can isolate the variable, $t$, by dividing the $2$ logarithms:

$\log3 = \log(1.064)^t$

$\log3 = t\log1.064$

$t = \cfrac{\log3}{\log1.064}$

$t \approx 17.7\;[\text{years}]$

Therefore, we can determine that it will take roughly $\boldsymbol{\approx 17.7\;[\textbf{years}]}$ for the original amount to triple.

A culture of bacteria triples every $30\;[\text{minutes}]$. How long will it take a culture originally consisting of $40 \; [\text{bacteria}]$ to grow to a population of $200 000 \; [\text{bacteria}]$.

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First, we can identify what each variable and value represents:

$B(m)$ represents the bacteria population that exits after each minute
$40$ represents the initial bacteria population
$3$ represents the Growth Factor
$m$ represents the period in minutes where the bacteria population increases
$30$ represents how often in days the bacteria population triples

Using all of the found values, we can write the equation as such:

$B(m) = 40(3)^{m/30}$

Next, we can set the equation equal to $200 000$. We can then divide both sides by $40$:

$200 000 = 40(3)^{m/30}$

$5000 = (3)^{m/30}$

Then, we can take the common log of both sides and use the Logarithm Power Rule to bring the exponent down. Then, we can isolate the variable, $t$, by dividing the $2$ logarithms:

$\log5000 = \log(3)^{m/30}$

$\log5000 = \cfrac{m}{30} \cdot \log3$

$\cfrac{m}{30} = \cfrac{\log5000}{\log3}$

$\cfrac{m}{30} = 7.752683344$

Finally, we can multiply both sides by $30$ to determine the total amount of time in minutes:

$m \approx 233\;[\text{minutes}]$

Therefore, we can determine that it will take $\boldsymbol{\approx 233\;[\textbf{minutes}]}$ for the initial bacteria population to grow to a size of $200 000$ bacteria.

A substance decays at a rate of $5\%$ per year. If the initial amount of the substance is $100$ grams, how many years will it take for the substance to decay to $50$ grams?

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First, we can identify what each variable and value represents:

$P(t)$ represents the substance amount after each year
$100$ represents the initial substance amount
$-0.05$ represents the Exponential Decay factor
$t$ represents the period in years where the substance amount decreases

Using all of the found values, we can write the equation as such:

$P(t) = 100e^{-0.05t}$

Next, we can set the equation equal to $50$. We can then divide both sides by $100$:

$50 = 100e^{-0.05t}$

$0.5 = e^{-0.05t}$

Then, we can take the Natural Logarithm of both sides:

$\ln0.5 = \ln e^{-0.05t}$

After, we can simplify the expression using the Property of Logarithms to bring down the exponent:

$-0.05t = \log0.5$

Finally, we can solve for $t$:

$t = \cfrac{\ln0.5}{-0.05}$

$t \approx 13. 9\;[\text{years}]$

Therefore, we can determine that it will take $\boldsymbol{\approx 13. 9\;[\textbf{years}]}$ for the substance to decay to $50 \; [\text{grams}]$.

Exponentials - Word Problems

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