IndietroComprehensive Calculus Study Guide: Step-by-Step Guidance for Exam 1
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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)
To solve $\ln y = 0$, recall that $y = 1$.
Step-by-Step Guidance
Combine the logarithms using the addition rule: $\ln x + \ln(x - 3) = \ln[x(x - 3)]$.
Set the combined logarithm equal to zero: $\ln[x(x - 3)] = 0$.
Recall that $\ln y = 0$ implies $y = 1$. Set $x(x - 3) = 1$.
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 solve equations involving logarithms and to use properties of logarithms to simplify and solve.
Key Terms and Formulas:
$\ln a + \ln b = \ln(ab)$
If $\ln y = \ln k$, then $y = k$.
Step-by-Step Guidance
Combine the logarithms: $\ln x + \ln(x - 3) = \ln[x(x - 3)]$.
Set the combined logarithm equal to $\ln 4$: $\ln[x(x - 3)] = \ln 4$.
Use the property that if $\ln y = \ln k$, then $y = k$. Set $x(x - 3) = 4$.
Write the resulting quadratic equation and prepare to solve for $x$.
Try solving on your own before revealing the answer!
Final Answer:
Solving $x(x - 3) = 4$ 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 expressing both sides with the same base and using logarithms.
Key Terms and Formulas:
Exponential property: $a^{b} = c$ can be solved by taking logarithms.
$\log_a c = b$
Step-by-Step Guidance
Express $21$ in terms of $7$ if possible, or take the natural logarithm of both sides.
Apply $\ln$ to both sides: $\ln(7^{2x+1}) = \ln(21)$.
Use the property $\ln(a^b) = b \ln a$ to rewrite the left side.
Set up the equation for $x$ and prepare to solve.
Try solving on your own before revealing the answer!
Final Answer:
$\ln(7^{2x+1}) = (2x+1)\ln 7 = \ln 21$
So $2x+1 = \frac{\ln 21}{\ln 7}$
$x = \frac{1}{2}\left(\frac{\ln 21}{\ln 7} - 1\right)$
Numerically, $\ln 21 / \ln 7 = \log_7 21 = 1.5$, so $x = \frac{1}{2}(1.5 - 1) = 0.25$
Final solution: $x = 0.25$
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:
$a^{b} = c \implies b = \log_a c$
$\ln(a^b) = b \ln a$
Step-by-Step Guidance
Take the natural logarithm of both sides: $\ln(3^{5x+2}) = \ln(12)$.
Rewrite the left side using $\ln(a^b) = b \ln a$.
Set up the equation for $x$ and prepare to solve.
Try solving on your own before revealing the answer!
Final Answer:
$\ln(3^{5x+2}) = (5x+2)\ln 3 = \ln 12$
$5x+2 = \frac{\ln 12}{\ln 3}$
$x = \frac{1}{5}\left(\frac{\ln 12}{\ln 3} - 2\right)$
Numerically, $\ln 12 / \ln 3 = \log_3 12 \approx 2.26186$, so $x \approx \frac{1}{5}(2.26186 - 2) \approx 0.052$
Final solution: $x \approx 0.052$
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:
$\ln x$ is the natural logarithm of $x$.
Exponentiate both sides to solve for $x$.
Step-by-Step Guidance
Divide both sides by 3 to isolate $\ln x$.
Exponentiate both sides to solve for $x$.
Try solving on your own before revealing the answer!
Final Answer:
$3\ln x = 5 \implies \ln x = \frac{5}{3}$
$x = e^{5/3}$
Final solution: $x = e^{5/3} \approx 5.294$
Q6. Solve the equation: $e^{2x^2 - 7x - 15} = 1$
Background
Topic: Exponential Equations
This question tests your ability to solve equations involving exponentials.
Key Terms and Formulas:
$e^y = 1$ implies $y = 0$
Step-by-Step Guidance
Recall that $e^y = 1$ only when $y = 0$.
Set $2x^2 - 7x - 15 = 0$.
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 $x = -1.5$
Final solutions: $x = 5$ and $x = -1.5$
