Math Problem Statement

Les fonctions définies par A(t) = 3000(1,16)^t/10 et B(t) = 5000(0,95)^t représentent les nombres d’élèves qui s’inscriront dans t années dans 2 collèges. Résoudre les questions a) à d).

Solution

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Question 11

The two functions A(t)A(t) and B(t)B(t) represent the number of students enrolling in two colleges after tt years:

  • A(t)=3000×(1.16)t/10A(t) = 3000 \times (1.16)^{t/10} for college A.
  • B(t)=5000×(0.95)tB(t) = 5000 \times (0.95)^t for college B.

a) Quel est le nombre d’élèves qui s’inscriront maintenant dans le collège A? We are tasked with finding the number of students currently enrolling in college A. To do this, we substitute t=0t = 0 (since "maintenant" indicates the current time): A(0)=3000×(1.16)0/10=3000×1=3000A(0) = 3000 \times (1.16)^{0/10} = 3000 \times 1 = 3000 Thus, 3000 students will enroll in college A now.

b) Après combien de temps le nombre d’inscriptions, pour le premier collège, sera-t-il de 3231? We need to find tt such that A(t)=3231A(t) = 3231: 3231=3000×(1.16)t/103231 = 3000 \times (1.16)^{t/10} Divide both sides by 3000: 32313000=(1.16)t/10\frac{3231}{3000} = (1.16)^{t/10} 1.077=(1.16)t/101.077 = (1.16)^{t/10} Now, take the natural logarithm of both sides: ln(1.077)=t10ln(1.16)\ln(1.077) = \frac{t}{10} \ln(1.16) Solve for tt: t=10×ln(1.077)ln(1.16)t = 10 \times \frac{\ln(1.077)}{\ln(1.16)}

c) Après combien de temps le deuxième collège (B) enregistrera-t-il 3317 inscriptions? We need to solve for tt such that B(t)=3317B(t) = 3317: 3317=5000×(0.95)t3317 = 5000 \times (0.95)^t Divide both sides by 5000: 33175000=(0.95)t\frac{3317}{5000} = (0.95)^t 0.6634=(0.95)t0.6634 = (0.95)^t Take the natural logarithm of both sides: ln(0.6634)=tln(0.95)\ln(0.6634) = t \ln(0.95) Solve for tt: t=ln(0.6634)ln(0.95)t = \frac{\ln(0.6634)}{\ln(0.95)}

d) Après combien de temps ces deux collèges recevront-ils le même nombre d’inscriptions? Set A(t)=B(t)A(t) = B(t) and solve for tt: 3000×(1.16)t/10=5000×(0.95)t3000 \times (1.16)^{t/10} = 5000 \times (0.95)^t This equation can be solved numerically to find the value of tt.


Question 12

The equation t=log1/6(h)t = \log_{1/6}(h) expresses the time elapsed in hours as a function of the height hh (in hundreds of meters) of a weather balloon.

a) À quel instant t le ballon est-il à une hauteur de 100 m? Substitute h=100h = 100 into the equation: t=log1/6(100)t = \log_{1/6}(100) This will require using logarithm properties and can be simplified with logarithmic conversion to base 10 or natural logarithms.

b) À quel instant t le ballon est-il à 100^{216} m de hauteur? For h=100216h = 100^{216}, plug this value into the equation: t=log1/6(100216)t = \log_{1/6}(100^{216}) Use logarithmic properties to simplify.

c) Trouve le domaine et l’image de t. The domain of tt is determined by the possible values of hh (height), which must be positive. The image (range) is the possible values for tt, which will depend on how hh behaves under the logarithmic transformation.

d) En théorie, est-il possible que le ballon touche le sol? For the balloon to touch the ground, the height hh must approach 0. We need to consider the limit of tt as hh approaches 0 and see if this leads to a finite or infinite time.


Would you like to go deeper into solving these problems, or do you have any questions regarding specific steps?

Here are some related questions:

  1. How can logarithmic properties help simplify the equations in questions 11 and 12?
  2. What is the interpretation of the base of a logarithmic function in a real-world scenario like this?
  3. How do you approach solving exponential equations involving different bases?
  4. Can we approximate the time for question 11c) using numerical methods?
  5. What are some common applications of exponential and logarithmic functions in real life?

Tip: Remember that solving logarithmic equations often requires changing the base or using logarithmic properties to simplify the expressions.

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Math Problem Analysis

Mathematical Concepts

Exponential Functions
Logarithmic Functions
Equation Solving

Formulas

A(t) = 3000 * (1.16)^(t/10)
B(t) = 5000 * (0.95)^t
Logarithmic properties for equation solving

Theorems

Properties of Exponents
Logarithmic Change of Base Theorem

Suitable Grade Level

Grades 11-12