Skip to main content
Indietro

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

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 how to solve equations involving natural logarithms.

Key Terms and Formulas

  • Natural logarithm: $\ln a$ is the logarithm base $e$ of $a$.

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

  • To solve $\ln y = 0$, recall that $e^0 = 1$, so $y = 1$.

Step-by-Step Guidance

  1. Combine the two logarithms using the property $\ln a + \ln b = \ln(ab)$.

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

  3. Exponentiate both sides to eliminate the logarithm: $x(x-3) = 1$.

  4. Rewrite the equation as a quadratic and set it equal to zero.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{3 \pm \sqrt{13}}{2}$

After combining and solving the quadratic $x^2 - 3x - 1 = 0$, use the quadratic formula to find the solutions. Only values where $x > 3$ are valid due to the domain of the logarithm.

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

Background

Topic: Exponential and Logarithmic Equations

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

Key Terms and Formulas

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

  • If $\ln y = \ln k$, then $y = k$ (as long as $y, k > 0$).

Step-by-Step Guidance

  1. Combine the left side using $\ln a + \ln b = \ln(ab)$ to get $\ln(x(x-3))$.

  2. Set $\ln(x(x-3)) = \ln 4$.

  3. Since the logarithms are equal, set their arguments equal: $x(x-3) = 4$.

  4. Rewrite as a quadratic equation and prepare to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{3 \pm \sqrt{25}}{2} = 4, -1$

Only $x=4$ is valid since $x>3$ for the domain of the logarithms. $x=-1$ is extraneous.

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

Background

Topic: Exponential Equations

This question tests your ability to solve equations where the variable is in the exponent.

Key Terms and Formulas

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

  • Take logarithms of both sides to bring down the exponent.

  • Use $\log_a b = \frac{\ln b}{\ln a}$ if needed.

Step-by-Step Guidance

  1. Rewrite $21$ as $7^1 \times 3$ or just 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 bring down the exponent.

  4. Isolate $x$ by solving the resulting linear equation.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{\ln 21 - \ln 7}{2 \ln 7}$

After simplifying, you can further reduce to $x = \frac{\ln 3}{2 \ln 7}$.

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

Background

Topic: Exponential Equations

This question tests your ability to solve for $x$ when it appears in the exponent.

Key Terms and Formulas

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

  • Take logarithms of both sides to solve for $x$.

Step-by-Step Guidance

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

  2. Use the property $\ln(a^b) = b\ln a$ to bring down the exponent.

  3. Expand and isolate $x$ in the resulting linear equation.

  4. Prepare to solve for $x$ by moving terms appropriately.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{\ln 12 - 2\ln 3}{5\ln 3}$

This comes from isolating $x$ after applying logarithms and simplifying.

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

  • Property: $a\ln x = \ln x^a$

  • To solve $\ln y = k$, exponentiate both sides: $y = e^k$

Step-by-Step Guidance

  1. Divide both sides by 3 to isolate $\ln x$.

  2. Exponentiate both sides to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $x = e^{5/3}$

After isolating $\ln x$, exponentiate both sides to get the solution.

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

Background

Topic: Exponential Equations

This question tests your understanding of the properties of the exponential function and how to solve for the exponent when the output is 1.

Key Terms and Formulas

  • Property: $e^k = 1$ if and only if $k = 0$

Step-by-Step Guidance

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

  2. Set the exponent equal to zero: $2x^2 - 7x - 15 = 0$.

  3. Rewrite as a quadratic equation and prepare to solve for $x$.

Try solving on your own before revealing the answer!

Final Answer: $x = \frac{7 \pm \sqrt{169}}{4} = 5.5, -1.5$

Solving the quadratic gives two real solutions for $x$.

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

Background

Topic: Logarithmic Properties

This question tests your ability to use properties of logarithms to combine and evaluate expressions.

Key Terms and Formulas

  • Property: $\log_b a + \log_b c = \log_b (ac)$

  • Evaluate $\log_b d$ by expressing $d$ as a power of $b$ if possible.

Step-by-Step Guidance

  1. Combine the two logarithms using $\log_b a + \log_b c = \log_b (ac)$.

  2. Multiply $9 \times 25$ to get the argument of the single logarithm.

  3. Express the result as $\log_{15} (225)$ and see if $225$ can be written as a power of $15$.

Try solving on your own before revealing the answer!

Final Answer: $\log_{15} 225 = 2$

Since $225 = 15^2$, 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 use properties of exponents and logarithms to simplify expressions.

Key Terms and Formulas

  • $e^{a\ln b} = b^a$

  • $e^{A+B} = e^A \cdot e^B$

Step-by-Step Guidance

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

  2. Combine the exponents: $e^{\ln 5^3 + \ln 2^7} = e^{\ln(5^3 \cdot 2^7)}$.

  3. Recall that $e^{\ln k} = k$ for $k > 0$.

  4. Multiply $5^3$ and $2^7$ to get the final value.

Try solving on your own before revealing the answer!

Final Answer: $e^{3\ln 5 + 7\ln 2} = 1000$

Because $5^3 = 125$ and $2^7 = 128$, $125 \times 128 = 16000$.

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 properties of logarithms to simplify and evaluate expressions with given values.

Key Terms and Formulas

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

  • $\log_b A^k = k \log_b A$

  • $\sqrt{x} = x^{1/2}$, $\sqrt[3]{z} = z^{1/3}$

Step-by-Step Guidance

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

  2. Apply the logarithm power rule: $\log_b x^{1/2} = \frac{1}{2} \log_b x$ and $\log_b z^{1/3} = \frac{1}{3} \log_b z$.

  3. Use the quotient rule: $\log_b \frac{\sqrt{x}}{\sqrt[3]{z}} = \frac{1}{2} \log_b x - \frac{1}{3} \log_b z$.

  4. Substitute the given values for $\log_b x$ and $\log_b z$.

Try solving on your own before revealing the answer!

Final Answer: $\frac{1}{2}(2.4) - \frac{1}{3}(3.6) = 1.2 - 1.2 = 0$

After substituting and simplifying, the value is $0$.

Pearson Logo

Study Prep