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

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Q1. Solve the equation: ln x + ln(x − 3) = 0

Background

Topic: Exponential and Logarithmic Equations

This question tests your ability to manipulate logarithmic expressions and solve equations involving logarithms.

Key Terms and Formulas:

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

  • $\ln a = 0 \implies a = 1$

Step-by-Step Guidance

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

  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:

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

Both values must be checked to ensure $x > 3$ for the logarithms to be defined. Only $x = \frac{3 + \sqrt{13}}{2}$ is valid.

Q2. Solve the equation: ln x + ln(x − 3) = ln 4

Background

Topic: Logarithmic Equations

This question tests your ability to solve equations involving logarithms and equate logarithmic expressions.

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 the logarithms to be defined.

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

Background

Topic: Exponential Equations

This question tests your ability to solve equations involving exponents by expressing both sides with the same base.

Key Terms and Formulas:

  • $a^{b} = c \implies b = \log_{a} c$

  • Properties of exponents: $a^{m+n} = a^{m} \cdot a^{n}$

Step-by-Step Guidance

  1. Express 21 in terms of base 7: $21 = 7 \times 3$.

  2. Rewrite the left side: $7^{2x+1} = 7^{2x} \cdot 7^1$.

  3. Set $7^{2x+1} = 7 \times 3$ and take logarithms to solve for $x$.

  4. Prepare to isolate $x$ using logarithmic properties.

Try solving on your own before revealing the answer!

Final Answer:

Take $\ln$ of both sides:

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

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

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

Numerically, $x \approx 0.23$

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

Background

Topic: Exponential Equations

This question tests your ability to solve exponential equations using logarithms.

Key Terms and Formulas:

  • $a^{b} = c \implies b = \log_{a} c$

  • Logarithm properties: $\ln(a^{b}) = b \ln a$

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 rewrite the left side.

  3. Set up the equation to solve for $x$.

  4. Prepare to isolate $x$.

Try solving on your own before revealing the answer!

Final Answer:

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

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

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

Numerically, $x \approx 0.43$

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

Background

Topic: Logarithmic Equations

This question tests your ability to solve for $x$ in equations involving logarithms.

Key Terms and Formulas:

  • $\ln x$ is the natural logarithm of $x$

  • Exponentiation: $x = e^{y}$ if $\ln x = y$

Step-by-Step Guidance

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

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

  3. Set up $x = e^{5/3}$.

Try solving on your own before revealing the answer!

Final Answer:

$x = e^{5/3}$

Numerically, $x \approx 5.294$

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

Background

Topic: Exponential Equations

This question tests your ability to solve equations where the exponent is a quadratic expression.

Key Terms and Formulas:

  • $e^{y} = 1 \implies y = 0$

Step-by-Step Guidance

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

  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:

Solving $2x^2 - 7x - 15 = 0$ gives:

$x = \frac{7 \pm \sqrt{49 + 120}}{4} = \frac{7 \pm \sqrt{169}}{4} = \frac{7 \pm 13}{4}$

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 logarithm 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 $9 \times 25$.

  3. Express the result as $\log_{15} (225)$.

  4. Prepare to simplify $\log_{15} (225)$.

Try solving on your own before revealing the answer!

Final Answer:

$\log_{15} 225 = 2$

Because $225 = 15^2$, so $\log_{15} 225 = \log_{15} 15^2 = 2$

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 exponentials and logarithms.

Key Terms and Formulas:

  • $e^{a \ln b} = b^{a}$

  • $e^{x+y} = e^{x} \cdot e^{y}$

Step-by-Step Guidance

  1. Rewrite $e^{3 \ln 5 + 7 \ln 2}$ as $e^{3 \ln 5} \cdot e^{7 \ln 2}$.

  2. Recall $e^{a \ln b} = b^{a}$, so $e^{3 \ln 5} = 5^3$ and $e^{7 \ln 2} = 2^7$.

  3. Multiply $5^3 \cdot 2^7$.

  4. Prepare to compute the final value.

Try solving on your own before revealing the answer!

Final Answer:

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

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

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 logarithm properties for roots and quotients.

Key Terms and Formulas:

  • $\log_{b} \frac{A}{B} = \log_{b} A - \log_{b} B$

  • $\log_{b} A^{r} = r \log_{b} A$

Step-by-Step Guidance

  1. Rewrite $\log_{b} \frac{\sqrt{x}}{\sqrt[3]{z}}$ as $\log_{b} \sqrt{x} - \log_{b} \sqrt[3]{z}$.

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

  3. Apply the exponent rule: $\log_{b} x^{1/2} = \frac{1}{2} \log_{b} x$, $\log_{b} z^{1/3} = \frac{1}{3} \log_{b} z$.

  4. Substitute the given values for $\log_{b} x$ and $\log_{b} z$.

  5. Set up the expression for the final calculation.

Try solving on your own before revealing the answer!

Final Answer:

$\log_{b} \frac{\sqrt{x}}{\sqrt[3]{z}} = \frac{1}{2} \cdot 2.4 - \frac{1}{3} \cdot 3.6 = 1.2 - 1.2 = 0$

The exact value is 0.

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