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 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
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 $\ln a = 0$ means $a = 1$. 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:
$x = \frac{3 \pm \sqrt{9 + 4}}{2} = \frac{3 \pm \sqrt{13}}{2}$
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
Combine the logarithms: $\ln x + \ln(x-3) = \ln[x(x-3)]$.
Set the combined logarithm equal to $\ln 4$: $\ln[x(x-3)] = \ln 4$.
Since $\ln a = \ln b$ implies $a = b$, set $x(x-3) = 4$.
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
Express 21 in terms of base 7: $21 = 7 \times 3$.
Rewrite the left side: $7^{2x+1} = 7^{2x} \cdot 7^1$.
Set $7^{2x+1} = 7 \times 3$ and take logarithms to solve for $x$.
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
Take the natural logarithm of both sides: $\ln(3^{5x+2}) = \ln(12)$.
Use the property $\ln(a^{b}) = b \ln a$ to rewrite the left side.
Set up the equation to solve for $x$.
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
Divide both sides by 3: $\ln x = \frac{5}{3}$.
Exponentiate both sides to solve for $x$.
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
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:
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
Combine the logarithms: $\log_{15} 9 + \log_{15} 25 = \log_{15} (9 \times 25)$.
Calculate $9 \times 25$.
Express the result as $\log_{15} (225)$.
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
Rewrite $e^{3 \ln 5 + 7 \ln 2}$ as $e^{3 \ln 5} \cdot e^{7 \ln 2}$.
Recall $e^{a \ln b} = b^{a}$, so $e^{3 \ln 5} = 5^3$ and $e^{7 \ln 2} = 2^7$.
Multiply $5^3 \cdot 2^7$.
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
Rewrite $\log_{b} \frac{\sqrt{x}}{\sqrt[3]{z}}$ as $\log_{b} \sqrt{x} - \log_{b} \sqrt[3]{z}$.
Recall $\sqrt{x} = x^{1/2}$ and $\sqrt[3]{z} = z^{1/3}$.
Apply the exponent rule: $\log_{b} x^{1/2} = \frac{1}{2} \log_{b} x$, $\log_{b} z^{1/3} = \frac{1}{3} \log_{b} z$.
Substitute the given values for $\log_{b} x$ and $\log_{b} z$.
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.