IndietroComprehensive Calculus I 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 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
Combine the two logarithms using the property $\ln a + \ln b = \ln(ab)$.
Set the resulting logarithm equal to zero: $\ln(x(x-3)) = 0$.
Exponentiate both sides to eliminate the logarithm: $x(x-3) = 1$.
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
Combine the left side using $\ln a + \ln b = \ln(ab)$ to get $\ln(x(x-3))$.
Set $\ln(x(x-3)) = \ln 4$.
Since the logarithms are equal, set their arguments equal: $x(x-3) = 4$.
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
Rewrite $21$ as $7^1 \times 3$ or just 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 bring down the exponent.
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
Take the natural logarithm of both sides: $\ln(3^{5x+2}) = \ln 12$.
Use the property $\ln(a^b) = b\ln a$ to bring down the exponent.
Expand and isolate $x$ in the resulting linear equation.
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
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: $x = e^{5/3}$
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
Recall that $e^y = 1$ only when $y = 0$.
Set the exponent equal to zero: $2x^2 - 7x - 15 = 0$.
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
Combine the two logarithms using $\log_b a + \log_b c = \log_b (ac)$.
Multiply $9 \times 25$ to get the argument of the single logarithm.
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
Rewrite $3\ln 5$ as $\ln 5^3$ and $7\ln 2$ as $\ln 2^7$.
Combine the exponents: $e^{\ln 5^3 + \ln 2^7} = e^{\ln(5^3 \cdot 2^7)}$.
Recall that $e^{\ln k} = k$ for $k > 0$.
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
Rewrite $\sqrt{x}$ as $x^{1/2}$ and $\sqrt[3]{z}$ as $z^{1/3}$.
Apply the logarithm power rule: $\log_b x^{1/2} = \frac{1}{2} \log_b x$ and $\log_b z^{1/3} = \frac{1}{3} \log_b z$.
Use the quotient rule: $\log_b \frac{\sqrt{x}}{\sqrt[3]{z}} = \frac{1}{2} \log_b x - \frac{1}{3} \log_b z$.
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 $0$.