Select the correct answer.
Insect larvae grow incredibly quickly over the first several days of their lives. The length of one particular species increases exponentially at a rate of $25\%$ each day. A scientist studying this species of larva measures one specimen to be $3 ~ mm$ in length at the beginning of his observation period. The scientist will transfer the larvae to a new environment once the larvae are over 9.15 millimeters long.
If $t$ represents the number of days since the scientist began his observations, write an inequality to represent the situation, and use it to determine after how many days the larvae are transferred to the new environment.
A. $t<6$
B. $t>9$
C. $t<3$
D. $t>5
Answer (2)
Write the exponential growth equation: L ( t ) = 3 ( 1.25 ) t .
Set up the inequality: 9.15"> 3 ( 1.25 ) t > 9.15 .
Solve for t : \frac{ln(3.05)}{ln(1.25)} \approx 4.997"> t > l n ( 1.25 ) l n ( 3.05 ) ≈ 4.997 .
Conclude that the larvae are transferred after more than 5 days: 5"> t > 5 . The answer is D .
Explanation
Problem Analysis Let's analyze the problem. We are given that the length of the larva increases exponentially at a rate of 25% each day. The initial length is 3 mm, and the larva is transferred when its length exceeds 9.15 mm. We need to find the number of days, t , after which the transfer occurs.
Exponential Growth Equation First, we write the equation for the length of the larva as a function of time t . Since the growth is exponential, we have: L ( t ) = 3 ( 1 + 0.25 ) t = 3 ( 1.25 ) t
Setting up the Inequality We want to find t such that 9.15"> L ( t ) > 9.15 . So, we set up the inequality: 9.15"> 3 ( 1.25 ) t > 9.15
Isolating the Exponential Term Now, we solve for t . Divide both sides by 3: \frac{9.15}{3} = 3.05"> ( 1.25 ) t > 3 9.15 = 3.05
Applying the Natural Logarithm To solve for t , we take the natural logarithm of both sides: ln(3.05)"> l n (( 1.25 ) t ) > l n ( 3.05 )
Using the Power Rule of Logarithms Using the power rule of logarithms, we have: ln(3.05)"> t l n ( 1.25 ) > l n ( 3.05 )
Solving for t Now, divide both sides by l n ( 1.25 ) :
\frac{ln(3.05)}{ln(1.25)}"> t > l n ( 1.25 ) l n ( 3.05 )
Calculating the Value of t We calculate the value of l n ( 1.25 ) l n ( 3.05 ) :
\frac{ln(3.05)}{ln(1.25)} \approx 4.997"> t > l n ( 1.25 ) l n ( 3.05 ) ≈ 4.997
Final Answer and Conclusion Since t must be greater than approximately 4.997, the smallest integer value for t that satisfies this condition is 5. Therefore, the larvae are transferred after more than 5 days. Looking at the options, the correct answer is: D. 5"> t > 5
Examples
Exponential growth is a common phenomenon in biology, finance, and other fields. For example, the growth of a bacteria colony, the accumulation of interest in a bank account, or the spread of a virus can all be modeled using exponential functions. Understanding exponential growth helps us predict future outcomes and make informed decisions. In this case, it helps the scientist determine when to transfer the larvae to a new environment to ensure their optimal growth.
The larvae grow exponentially at a rate of 25% per day, starting at 3 mm and needing to exceed 9.15 mm for transfer. The inequality to set up is 9.15"> 3 ( 1.25 ) t > 9.15 , which leads to determining that 5"> t > 5 . Thus, the correct answer is D: 5"> t > 5 .
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