IndietroComprehensive 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 understanding of logarithmic properties and solving equations involving logarithms.
Key Terms and Formulas:
$\ln a + \ln b = \ln(ab)$ (Logarithm addition property)
To solve $\ln y = 0$, recall that $y = 1$.
Step-by-Step Guidance
Combine the logarithms using the property: $\ln x + \ln(x - 3) = \ln[x(x - 3)]$.
Set the combined logarithm equal to zero: $\ln[x(x - 3)] = 0$.
Recall that $\ln y = 0$ means $y = 1$. Set $x(x - 3) = 1$.
Write the resulting quadratic equation: $x^2 - 3x - 1 = 0$.
Try solving on your own before revealing the answer!
Final Answer:
Solving $x^2 - 3x - 1 = 0$ gives $x = \frac{3 \pm \sqrt{13}}{2}$.
Check that $x > 3$ for the logarithms to be defined. Only $x = \frac{3 + \sqrt{13}}{2}$ is valid.
The solution is $x = \frac{3 + \sqrt{13}}{2}$.
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 $x$.
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 equal to $\ln 4$: $\ln[x(x - 3)] = \ln 4$.
Since $\ln y = \ln k$ implies $y = k$, set $x(x - 3) = 4$.
Write the quadratic equation: $x^2 - 3x - 4 = 0$.
Try solving on your own before revealing the answer!
Final Answer:
Solving $x^2 - 3x - 4 = 0$ gives $x = 4$ and $x = -1$.
Check domain: $x > 3$ for $x - 3 > 0$. Only $x = 4$ is valid.
The solution is $x = 4$.
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 equation: $a^{bx+c} = d$
Take logarithms to solve for $x$.
Step-by-Step Guidance
Rewrite $21$ as $7^1 \times 3$ to see if bases can be matched, 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 get $(2x+1)\ln 7 = \ln 21$.
Isolate $x$ by rearranging: $2x+1 = \frac{\ln 21}{\ln 7}$.
Try solving on your own before revealing the answer!
Final Answer:
$x = \frac{1}{2}\left(\frac{\ln 21}{\ln 7} - 1\right)$
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:
Exponential equation: $a^{bx+c} = d$
Take logarithms to solve for $x$.
Step-by-Step Guidance
Take the natural logarithm of both sides: $\ln(3^{5x+2}) = \ln(12)$.
Apply $\ln(a^b) = b \ln a$ to get $(5x+2)\ln 3 = \ln 12$.
Isolate $x$: $5x+2 = \frac{\ln 12}{\ln 3}$.
Solve for $x$ by rearranging: $x = \frac{1}{5}\left(\frac{\ln 12}{\ln 3} - 2\right)$.
Try solving on your own before revealing the answer!
Final Answer:
$x = \frac{1}{5}\left(\frac{\ln 12}{\ln 3} - 2\right)$
Plug in values to get $x \approx 0.46$.
Q5. Solve the equation: $3\ln x = 5$
Background
Topic: Logarithmic Equations
This question tests your ability to solve for $x$ using properties of logarithms and exponentials.
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: $\ln x = \frac{5}{3}$.
Exponentiate both sides: $x = e^{\frac{5}{3}}$.
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 $x$ when the exponent of $e$ equals zero.
Key Terms and Formulas:
$e^y = 1$ if and only if $y = 0$.
Step-by-Step Guidance
Set the exponent equal to zero: $2x^2 - 7x - 15 = 0$.
This is a quadratic equation; use the quadratic formula to solve for $x$.
Quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -7$, $c = -15$.
Try solving on your own before revealing the answer!
Final Answer:
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 logarithm addition properties and evaluate logarithms.
Key Terms and Formulas:
$\log_a b + \log_a c = \log_a (bc)$
Step-by-Step Guidance
Combine the logs: $\log_{15} 9 + \log_{15} 25 = \log_{15} (9 \times 25)$.
Calculate $9 \times 25 = 225$.
Express $225$ as a power of $15$ if possible.
Try solving on your own before revealing the answer!
Final Answer:
$225 = 15^2$, so $\log_{15} 225 = \log_{15} 15^2 = 2$.
The exact value 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 using properties of logarithms and exponents.
Key Terms and Formulas:
$e^{a \ln b} = b^a$
$e^{\ln b + \ln c} = bc$
Step-by-Step Guidance
Rewrite $3 \ln 5$ as $\ln 5^3$ and $7 \ln 2$ as $\ln 2^7$.
Add the logarithms: $\ln 5^3 + \ln 2^7 = \ln(5^3 \times 2^7)$.
Exponentiate: $e^{\ln(5^3 \times 2^7)} = 5^3 \times 2^7$.
Try solving on your own before revealing the answer!
Final Answer:
$5^3 = 125$, $2^7 = 128$, so $125 \times 128 = 16,000$.
The exact value is $16,000$.
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 properties of logarithms to simplify and evaluate expressions.
Key Terms and Formulas:
$\log_b \sqrt{x} = \frac{1}{2} \log_b x$
$\log_b \sqrt[3]{z} = \frac{1}{3} \log_b z$
$\log_b \frac{a}{b} = \log_b a - \log_b b$
Step-by-Step Guidance
Apply the root properties: $\log_b \sqrt{x} = \frac{1}{2} \log_b x$, $\log_b \sqrt[3]{z} = \frac{1}{3} \log_b z$.
Apply the quotient property: $\log_b \frac{\sqrt{x}}{\sqrt[3]{z}} = \log_b \sqrt{x} - \log_b \sqrt[3]{z}$.
Substitute the given values: $\frac{1}{2} \times 2.4 - \frac{1}{3} \times 3.6$.
Try solving on your own before revealing the answer!
Final Answer:
$\frac{1}{2} \times 2.4 = 1.2$, $\frac{1}{3} \times 3.6 = 1.2$, so $1.2 - 1.2 = 0$.
The exact value is $0$.