Skip to main content
Indietro

Comprehensive Calculus Exam 1 Study Guide: Step-by-Step Guidance

NotePraticaVideolezioni

Guida di studio - Note intelligenti

Appunti personalizzati basati sui tuoi materiali, ampliati con definizioni chiave, esempi e contesto.

Q1. Solve the equation: $\ln x + \ln(x - 3) = 0$

Background

Topic: Exponential and Logarithmic Equations

This question tests your understanding of logarithmic properties and solving equations involving logarithms.

Key Terms and Formulas:

  • $\ln a + \ln b = \ln(ab)$ (Logarithm addition property)

  • $\ln a = 0 \implies a = 1$

Step-by-Step Guidance

  1. Combine the logarithms using the addition property: $\ln x + \ln(x - 3) = \ln[x(x - 3)]$.

  2. Set the combined logarithm equal to zero: $\ln[x(x - 3)] = 0$.

  3. Recall that $\ln a = 0$ means $a = 1$. Set $x(x - 3) = 1$.

  4. Write the resulting quadratic equation and prepare to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer:

Solving $x(x - 3) = 1$ gives $x^2 - 3x - 1 = 0$. Using the quadratic formula:

$x = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$

Check that both solutions are valid (i.e., $x > 3$ for $\ln(x - 3)$ to be defined). Only $x = \frac{3 + \sqrt{13}}{2}$ is valid.

Final solution: $x = \frac{3 + \sqrt{13}}{2}$

Q2. Solve the equation: $\ln x + \ln(x - 3) = \ln 4$

Background

Topic: Logarithmic Equations

This question tests your ability to manipulate logarithmic expressions and solve for the variable.

Key Terms and Formulas:

  • $\ln a + \ln b = \ln(ab)$

  • If $\ln a = \ln b$, then $a = b$

Step-by-Step Guidance

  1. Combine the logarithms: $\ln x + \ln(x - 3) = \ln[x(x - 3)]$.

  2. Set the combined logarithm equal to $\ln 4$: $\ln[x(x - 3)] = \ln 4$.

  3. Since $\ln a = \ln b$ implies $a = b$, set $x(x - 3) = 4$.

  4. Write the quadratic equation and prepare to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer:

Solving $x(x - 3) = 4$ gives $x^2 - 3x - 4 = 0$. Using the quadratic formula:

$x = \frac{3 \pm \sqrt{9 + 16}}{2} = \frac{3 \pm 5}{2}$

So $x = 4$ or $x = -1$. Only $x = 4$ is valid (since $x > 3$ for $\ln(x - 3)$ to be defined).

Final solution: $x = 4$

Q3. Solve the equation: $7^{2x+1} = 21$

Background

Topic: Exponential Equations

This question tests your ability to solve equations involving exponents by using logarithms.

Key Terms and Formulas:

  • Exponential property: $a^{bx+c} = d$

  • Take logarithms to solve for $x$

Step-by-Step Guidance

  1. Rewrite $21$ as $7^1 \times 3$ to see if bases can be matched, or take the natural logarithm of both sides.

  2. Apply $\ln$ to both sides: $\ln(7^{2x+1}) = \ln(21)$.

  3. Use the property $\ln(a^b) = b \ln a$ to simplify: $(2x+1)\ln 7 = \ln 21$.

  4. Isolate $x$ and prepare to solve.

Try solving on your own before revealing the answer!

Final Answer:

$(2x+1)\ln 7 = \ln 21$

$2x+1 = \frac{\ln 21}{\ln 7}$

$x = \frac{1}{2}\left(\frac{\ln 21}{\ln 7} - 1\right)$

Numerically, $\ln 21 / \ln 7 = \ln(7^1 \times 3) / \ln 7 = 1 + \ln 3 / \ln 7$; plug in values to get $x \approx 0.23$.

Q4. Solve the equation: $3^{5x+2} = 12$

Background

Topic: Exponential Equations

This question tests your ability to solve for $x$ in an exponential equation using logarithms.

Key Terms and Formulas:

  • $a^{bx+c} = d$

  • Take logarithms to solve for $x$

Step-by-Step Guidance

  1. Take the natural logarithm of both sides: $\ln(3^{5x+2}) = \ln(12)$.

  2. Use $\ln(a^b) = b \ln a$ to simplify: $(5x+2)\ln 3 = \ln 12$.

  3. Isolate $x$ and prepare to solve.

Try solving on your own before revealing the answer!

Final Answer:

$(5x+2)\ln 3 = \ln 12$

$5x+2 = \frac{\ln 12}{\ln 3}$

$x = \frac{1}{5}\left(\frac{\ln 12}{\ln 3} - 2\right)$

Numerically, $\ln 12 / \ln 3 \approx 2.26$, so $x \approx 0.05$.

Q5. Solve the equation: $3\ln x = 5$

Background

Topic: Logarithmic Equations

This question tests your ability to solve for $x$ in a logarithmic equation.

Key Terms and Formulas:

  • $a \ln x = b \implies \ln x = \frac{b}{a}$

  • $\ln x = c \implies x = e^c$

Step-by-Step Guidance

  1. Divide both sides by 3: $\ln x = \frac{5}{3}$.

  2. Exponentiate both sides to solve for $x$: $x = e^{5/3}$.

Try solving on your own before revealing the answer!

Final Answer:

$x = e^{5/3} \approx 5.294$

This is the exact value; you can use a calculator for the decimal approximation.

Q6. Solve the equation: $e^{2x^2 - 7x - 15} = 1$

Background

Topic: Exponential Equations

This question tests your ability to solve for $x$ in an exponential equation.

Key Terms and Formulas:

  • $e^a = 1 \implies a = 0$

Step-by-Step Guidance

  1. Recall that $e^a = 1$ only when $a = 0$.

  2. Set $2x^2 - 7x - 15 = 0$.

  3. Write the quadratic equation and prepare to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer:

$2x^2 - 7x - 15 = 0$

Using the quadratic formula: $x = \frac{7 \pm \sqrt{49 + 120}}{4} = \frac{7 \pm \sqrt{169}}{4} = \frac{7 \pm 13}{4}$

So $x = 5$ or $x = -1.5$

Q7. Find the exact value: $\log_{15} 9 + \log_{15} 25$

Background

Topic: Logarithmic Properties

This question tests your ability to use logarithmic addition properties and evaluate logarithms.

Key Terms and Formulas:

  • $\log_a b + \log_a c = \log_a (bc)$

Step-by-Step Guidance

  1. Combine the logarithms: $\log_{15} 9 + \log_{15} 25 = \log_{15} (9 \times 25)$.

  2. Calculate $9 \times 25 = 225$.

  3. Express $225$ as a power of $15$ if possible.

Try solving on your own before revealing the answer!

Final Answer:

$\log_{15} 225 = \log_{15} (15^2) = 2$

Because $225 = 15^2$, the answer is $2$.

Q8. Find the exact value: $e^{3 \ln 5 + 7 \ln 2}$

Background

Topic: Exponential and Logarithmic Properties

This question tests your ability to simplify expressions involving exponents and logarithms.

Key Terms and Formulas:

  • $e^{a \ln b} = b^a$

  • $e^{\ln a + \ln b} = ab$

Step-by-Step Guidance

  1. Rewrite $3 \ln 5 + 7 \ln 2$ as $\ln 5^3 + \ln 2^7$.

  2. Combine using $\ln a + \ln b = \ln(ab)$: $\ln(5^3 \times 2^7)$.

  3. Exponentiate: $e^{\ln(5^3 \times 2^7)} = 5^3 \times 2^7$.

Try solving on your own before revealing the answer!

Final Answer:

$5^3 = 125$, $2^7 = 128$

$5^3 \times 2^7 = 125 \times 128 = 16,000$

So $e^{3 \ln 5 + 7 \ln 2} = 16,000$

Q9. Find the exact value: $\log_b \frac{\sqrt{x}}{\sqrt[3]{z}}$ when $\log_b x = 2.4$ and $\log_b z = 3.6$

Background

Topic: Logarithmic Properties

This question tests your ability to use logarithmic properties to simplify and evaluate expressions.

Key Terms and Formulas:

  • $\log_b \frac{A}{B} = \log_b A - \log_b B$

  • $\log_b x^r = r \log_b x$

Step-by-Step Guidance

  1. Rewrite $\log_b \frac{\sqrt{x}}{\sqrt[3]{z}}$ as $\log_b x^{1/2} - \log_b z^{1/3}$.

  2. Apply the power rule: $\frac{1}{2} \log_b x - \frac{1}{3} \log_b z$.

  3. Substitute the given values: $\log_b x = 2.4$, $\log_b z = 3.6$.

Try solving on your own before revealing the answer!

Final Answer:

$\frac{1}{2} \times 2.4 - \frac{1}{3} \times 3.6 = 1.2 - 1.2 = 0$

The exact value is $0$.

Pearson Logo

Study Prep