Problem 35

Question

In Exercises \(34-39,\) solve the inequality analytically. $$ 1000(1.005)^{12 t} \geq 3000 $$

Step-by-Step Solution

Verified
Answer
\( t \geq 18.36 \)
1Step 1: Isolate the Exponential Term
Start by dividing both sides of the inequality by 1000 to isolate the exponential term. This gives:\[ (1.005)^{12t} \geq 3 \]
2Step 2: Apply the Logarithm
To solve for \( t \), take the natural logarithm on both sides of the inequality. This gives:\[ \ln((1.005)^{12t}) \geq \ln(3) \]
3Step 3: Use the Power Rule for Logarithms
Apply the power rule for logarithms, which states \( \ln(a^b) = b \ln(a) \). Thus,\[ 12t \ln(1.005) \geq \ln(3) \]
4Step 4: Solve for \( t \)
Divide both sides by \( 12 \ln(1.005) \) to solve for \( t \):\[ t \geq \frac{\ln(3)}{12 \ln(1.005)} \]
5Step 5: Calculate the Value of \( t \)
Now, compute the numerical value for \( t \) using a calculator:\[ \ln(3) \approx 1.0986 \] and \[ \ln(1.005) \approx 0.004987 \]. Thus,\[ t \geq \frac{1.0986}{12 \times 0.004987} \approx \frac{1.0986}{0.059844} \approx 18.36 \]

Key Concepts

Solving InequalitiesLogarithm PropertiesMathematical Problem-Solving
Solving Inequalities
Inequalities are a crucial part of mathematics and involve finding the values of variables that make an inequality true. Solving exponential inequalities involves isolating the exponential term and then using properties of logarithms to simplify and solve the inequality.
  • First, simplify the inequality by isolating the exponential expression. This can often be done by dividing both sides by the same positive number.
  • For exponential inequalities, taking the logarithm of both sides can help simplify the expression. This is because logarithms allow you to "bring down" the exponent as a coefficient, which makes solving for the variable easier.
  • Once the variable is isolated, solve like a regular inequality, paying attention to any conditions that might affect the direction of the inequality, such as dividing or multiplying by a negative number.
Remember to verify your result to ensure it satisfies the original inequality, maintaining a clear pathway from the original equation to your solution.
Logarithm Properties
Logarithms are incredibly useful in solving equations where the variable is in an exponent, such as exponential inequalities. Their properties allow us to simplify complex exponential terms into linear ones.
  • The most useful property is the power rule: \( \ln(a^b) = b \ln(a) \). This allows you to break down an exponential term into a simpler multiplication problem.
  • The product rule \( \ln(xy) = \ln(x) + \ln(y) \) and the quotient rule \( \ln\left(\frac{x}{y}\right) = \ln(x) - \ln(y) \) help in breaking down more complex logarithmic expressions.
  • Understanding how to switch between exponential and logarithmic forms can unlock solutions that may not be immediately obvious. For example, if \( x = a^y \), taking the logarithm gives \( \ln(x) = y \ln(a) \), which is easier to solve for \( y \).
These properties make logarithms a powerful tool in your mathematical toolkit.
Mathematical Problem-Solving
Approaching mathematical problems, like solving exponential inequalities, involves a systematic and strategic process. Here are some tips to improve your problem-solving skills:
  • **Understand the problem:** Carefully read the problem to understand what it is asking. Identify the goal and the given information.
  • **Plan your approach:** Consider the methods that can be employed, including any relevant mathematical principles, theories, or shortcuts.
  • **Carry out your plan:** Execute the steps carefully, checking your work as you go to ensure accuracy.
  • **Review the solution:** After solving, review your steps and the solution to confirm that the results meet the conditions of the original problem.
Problem-solving in mathematics is not just about getting to the answer, but also about understanding the process and modeling similar strategies to solve different types of problems. This habit builds a strong foundation for tackling more complex mathematical challenges.