Example Question - natural logarithm
Here are examples of questions we've helped users solve.
Combining Logarithmic Expressions into a Single Logarithm
<p>The expression given is \( \ln(7) + 4\ln(4) \).</p> <p>First, use the logarithm power rule \( a\ln(b) = \ln(b^a) \) to rewrite \( 4\ln(4) \) as \( \ln(4^4) \).</p> <p>Thus, \( 4\ln(4) = \ln(4^4) = \ln(256) \).</p> <p>Now, use the logarithm addition rule \( \ln(a) + \ln(b) = \ln(a \cdot b) \) to combine \( \ln(7) \) and \( \ln(256) \) into a single logarithm.</p> <p>\( \ln(7) + \ln(256) = \ln(7 \cdot 256) = \ln(1792) \).</p> <p>Therefore, \( \ln(7) + 4\ln(4) \) can be expressed as a single logarithm \( \ln(1792) \).</p>
Integral of a Rational Function with a Substitution
Para resolver la integral que has proporcionado, debemos enfocarnos en el integrando: \[\int \frac{3x^2 - 2}{x^3 + x^2 - 2x} dx\] Primero, se debe factorizar el denominador para poder descomponer la fracción en fracciones parciales si es necesario. El denominador factoriza como: \[x^3 + x^2 - 2x = x(x^2 + x - 2) = x(x+2)(x-1)\] Ahora podemos intentar descomponer la fracción original en fracciones parciales. Sin embargo, en este caso se puede observar que el numerador es la derivada del denominador, lo cual sugiere que una sustitución directa podría ser útil. Si dejamos que \( u = x^3 + x^2 - 2x \), entonces \( du = (3x^2 + 2x - 2)dx \), que es casi el numerador que tenemos. Hay un \(2x\) de más que podemos arreglar sustrayendo y sumando \(2x\) dentro del integrando: \[\int \frac{3x^2 - 2 + 2x - 2x}{x(x+2)(x-1)} dx = \int \frac{(3x^2 + 2x - 2) - 2x}{x(x+2)(x-1)} dx\] Esto nos da dos integrales separadas para resolver: \[= \int \frac{3x^2 + 2x - 2}{x(x+2)(x-1)} dx - \int \frac{2x}{x(x+2)(x-1)} dx \] La primera integral ahora es directa con la sustitución \(u = x^3 + x^2 - 2x\), resultando en: \[= \int \frac{1}{u} du - \int \frac{2x}{x(x+2)(x-1)} dx \] La primera integral se resuelve como el logaritmo natural: \[= \ln|u| - \int \frac{2x}{x(x+2)(x-1)} dx \] La segunda integral todavía necesita ser descompuesta en fracciones parciales, pero antes simplificamos cancelando el término x: \[= \ln|x^3 + x^2 - 2x| - \int \frac{2}{(x+2)(x-1)} dx \] Ahora se descompone la fracción \(\frac{2}{(x+2)(x-1)}\) en fracciones parciales. Para eso, proponemos que: \[\frac{2}{(x+2)(x-1)} = \frac{A}{x+2} + \frac{B}{x-1}\] Multiplicando ambos lados por el denominador común, \( (x+2)(x-1) \), y luego igualando los coeficientes, encontramos los valores de A y B. Después de eso, podemos integrar término por término. No proseguiré con esos cálculos aquí para mantener la explicación más breve. Al final, la solución a la integral original será una combinación del logaritmo natural de \(u\) y las integrales de las fracciones parciales que en este caso serán también logaritmos naturales.
Solving Exponential Equation with Logarithms
The equation given in the image is: 5^(3-x) + 2 = 8 To solve for x, we should first isolate the exponential term on one side of the equation: 5^(3-x) + 2 - 2 = 8 - 2 5^(3-x) = 6 Now, to solve for x, we need to rewrite the equation in logarithmic form. Taking the logarithm of both sides (using a common logarithm, base 10, or a natural logarithm, base e), we can apply the properties of logarithms to solve for the exponent. But first, let's use the natural logarithm, as it's often the one used in these types of problems: ln(5^(3-x)) = ln(6) Applying the property of logarithms that allows us to bring down the exponent, we get: (3 - x)ln(5) = ln(6) Now, divide both sides by ln(5) to isolate (3 - x): 3 - x = ln(6) / ln(5) Now, solve for x: x = 3 - ln(6) / ln(5) Using a calculator to find the values of ln(6) and ln(5) and then performing the subtraction: x ≈ 3 - (1.79176 / 1.60944) x ≈ 3 - 1.1134 x ≈ 1.8866 Rounded to the nearest thousandth, the value of x is approximately: x ≈ 1.887
Evaluating Logarithmic Expressions
To solve this problem, let's evaluate each logarithmic expression and see if it has been evaluated correctly. A. ln(0) = 1 This is incorrect because the natural logarithm of zero is undefined. ln(0) does not have a real value since you cannot raise e (Euler's number, approximately 2.71828) to any power to get zero. B. log₂(8) = 3 This is correct because 2 raised to the power 3 equals 8. (2^3 = 8) C. log(0.1) = -1 This is correct because 10 raised to the power -1 equals 0.1. (10^-1 = 0.1) D. log₂(-1) = 0.5 This is incorrect because you cannot take the logarithm of a negative number in the real number system. log₂(-1) is undefined. E. log(0.025) = -2 This is incorrect because 10 raised to the power of -2 equals 0.01, not 0.025. We need to find the actual value of log(0.025) to see if it equals -2. Let's calculate the actual value of log(0.025), which means we are looking for the exponent that 10 must be raised to in order to get 0.025. Using a calculator, log(0.025) ≈ -1.60206 Therefore, the options B and C have been evaluated correctly.
Correct Evaluation of Logarithmic Expressions
To solve this question, we need to evaluate the given logarithmic expressions and see which ones are correct. Here's an evaluation of each option: A. \(\ln(1) = 0\) This statement is true. The natural logarithm of 1 is always 0 because \(e^0 = 1\). B. \(\log_2 9 = 3\) This statement is false. The base 2 logarithm of 9 is not 3, because \(2^3 = 8\), not 9. C. \(\log_{10} \frac{1}{100} = \frac{1}{2}\) This statement is false. The base 10 logarithm of \(\frac{1}{100}\) is actually -2, because \(10^{-2} = \frac{1}{100}\). D. \(\log_3 (-1) = \frac{1}{3}\) This statement is false. The logarithm of a negative number is not a real number, and therefore, this expression cannot be evaluated in the real number system. E. \(\log_5 \frac{1}{125} = -3\) This statement is true. The base 5 logarithm of \(\frac{1}{125}\) is indeed -3, because \(5^{-3} = \frac{1}{125}\). So, the correctly evaluated logarithmic expressions are A and E.
Estimating the Natural Logarithm of 16
The question is asking for the best estimate of the natural logarithm of 16. To estimate the value without a calculator, it helps to know the value of the natural logarithm for some key numbers. For example: - ln(1) = 0 - ln(e) = 1, where e is approximately equal to 2.718 The value of ln(16) will be somewhere between the ln(e^2) and ln(e^3) since 16 is between e^2 and e^3. Knowing that: - e^2 is approximately 7.389 (since 2.718 squared is about 7.389) - e^3 is approximately 20.086 (since 2.718 cubed is about 20.086) We can tell that ln(16) will be close to but less than 3 because 16 is less than e^3 but greater than e^2. Among the choices given, the best estimate for ln(16) which is just a bit under 3, would be: C. 3.23 This is because 3.23 is the closest value under 3, which makes sense for an estimate of ln(16). It's a rough estimate, but it's the best fit given the options provided.
Derivative Calculation using Logarithmic Differentiation
To find the derivative f'(1) of the function f(x) using logarithmic differentiation, we start by taking the natural logarithm of both sides of the function. Given: f(x) = (2x - 1)^15 * (5x - 4)^5 * (9x - 8)^5 / (11x - 10)^2 * (14x - 13)^3 * (17x - 16)^3 Let y = f(x), then ln(y) = ln(f(x)) Taking logarithms: ln(y) = 15 * ln(2x - 1) + 5 * ln(5x - 4) + 5 * ln(9x - 8) - 2 * ln(11x - 10) - 3 * ln(14x - 13) - 3 * ln(17x - 16) Now, differentiate both sides with respect to x: 1/y * dy/dx = 15/(2x - 1) * 2 + 5/(5x - 4) * 5 + 5/(9x - 8) * 9 - 2/(11x - 10) * 11 - 3/(14x - 13) * 14 - 3/(17x - 16) * 17 Simplify: dy/dx = y * [ 30/(2x - 1) + 25/(5x - 4) + 45/(9x - 8) - 22/(11x - 10) - 42/(14x - 13) - 51/(17x - 16) ] Now we need to substitute x = 1 into the equation to find f'(1). First, evaluate y when x = 1, which is f(1), to substitute into the equation for dy/dx. f(1) = (2*1 - 1)^15 * (5*1 - 4)^5 * (9*1 - 8)^5 / (11*1 - 10)^2 * (14*1 - 13)^3 * (17*1 - 16)^3 f(1) = (1)^15 * (1)^5 * (1)^5 / (1)^2 * (1)^3 * (1)^3 f(1) = 1 Now we substitute x = 1 and f(1) = 1 into the derivative equation: f'(1) = 1 * [ 30/(2*1 - 1) + 25/(5*1 - 4) + 45/(9*1 - 8) - 22/(11*1 - 10) - 42/(14*1 - 13) - 51/(17*1 - 16) ] f'(1) = [ 30/1 + 25/1 + 45/1 - 22/1 - 42/1 - 51/1 ] f'(1) = 30 + 25 + 45 - 22 - 42 - 51 f'(1)= 35 Therefore, using logarithmic differentiation, we find that f'(1) = 35.
Derivative Calculation Using Logarithmic Differentiation
To differentiate the given function \( f(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} \) using logarithmic differentiation, we first take the natural logarithm of both sides of the equation defining \( f(x) \): 1. Take the natural logarithm of f(x): \[ \ln(f(x)) = \ln\left(\frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7}\right) \] 2. Apply properties of logarithms to simplify the right-hand side: \[ \ln(f(x)) = \ln(x^3) + \ln((3 - 2x)^2) - \ln(\sqrt[3]{x} + 7) \] \[ \ln(f(x)) = 3\ln(x) + 2\ln(3 - 2x) - \ln(\sqrt[3]{x} + 7) \] 3. Differentiate both sides with respect to x: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{d}{dx}(3\ln(x)) + \frac{d}{dx}(2\ln(3 - 2x)) - \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) \] 4. Apply the derivatives; - For \( \frac{d}{dx}(3\ln(x)) \), use the derivative of \(\ln(x)\), which is \( \frac{1}{x} \): \[ \frac{d}{dx}(3\ln(x)) = 3 \cdot \frac{1}{x} = \frac{3}{x} \] - For \( \frac{d}{dx}(2\ln(3 - 2x)) \), use the chain rule: \[ \frac{d}{dx}(2\ln(3 - 2x)) = 2 \cdot \frac{1}{(3 - 2x)} \cdot (-2) = \frac{-4}{(3 - 2x)} \] - For \( \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) \), again use the chain rule: \[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{d}{dx}(\sqrt[3]{x}) \] \[ \frac{d}{dx}(\sqrt[3]{x}) = \frac{1}{3x^{2/3}} \] So, \[ \frac{d}{dx}(\ln(\sqrt[3]{x} + 7)) = \frac{1}{(\sqrt[3]{x} + 7)} \cdot \frac{1}{3x^{2/3}} \] 5. Now combine these results: \[ \frac{1}{f(x)} \cdot f'(x) = \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \] 6. Solve for \( f'(x) \): \[ f'(x) = f(x) \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \] 7. Replace \( f(x) \) with the original function: \[ f'(x) = \frac{x^3 (3 - 2x)^2}{\sqrt[3]{x} + 7} \cdot \left( \frac{3}{x} - \frac{4}{3 - 2x} - \frac{1}{3x^{2/3}(\sqrt[3]{x} + 7)} \right) \] This is the derivative of the function using logarithmic differentiation. You can simplify the expression further by combining terms and multiplying through by the original function.
Solving Exponential Equation with Natural Logarithm
To solve the equation \( e^{4 - 7x} + 11 = 20 \), you first isolate the exponential part: 1. Subtract 11 from both sides to move the constant term to the right-hand side of the equation: \[ e^{4 - 7x} = 20 - 11 \] \[ e^{4 - 7x} = 9 \] 2. Now take the natural logarithm (ln) of both sides to solve for \( 4 - 7x \): \[ \ln(e^{4 - 7x}) = \ln(9) \] Since the natural logarithm and the exponential function are inverse functions, \( \ln(e^{y}) = y \), you get: \[ 4 - 7x = \ln(9) \] 3. Finally, solve for x: \[ 7x = 4 - \ln(9) \] \[ x = \frac{4 - \ln(9)}{7} \] This is the solution to the equation.
Solving Exponential Equation for x
The equation provided in the image is: \[ e^{4-7x} + 11 = 20 \] To solve for \( x \), we'll follow these steps: 1. Subtract 11 from both sides of the equation. \[ e^{4-7x} = 9 \] 2. Take the natural logarithm (ln) of both sides to get rid of the base \( e \). \[ \ln(e^{4-7x}) = \ln(9) \] 3. Use the property of logarithms that \( \ln(e^y) = y \) to simplify the left side. \[ 4 - 7x = \ln(9) \] 4. Subtract 4 from both sides to isolate the term with \( x \). \[ -7x = \ln(9) - 4 \] 5. Finally, divide both sides by -7 to solve for \( x \). \[ x = \frac{\ln(9) - 4}{-7} \] 6. Now you can either leave the answer in terms of natural logarithm or use a calculator to find a numerical value for \( x \). Using a calculator: \[ x \approx \frac{2.1972 - 4}{-7} \] \[ x \approx \frac{-1.8028}{-7} \] \[ x \approx 0.25754 \] So the solution to the equation is \( x \approx 0.25754 \).
Solving Exponential Equation for x
The equation provided is: \[ e^{4 - 7x} + 11 = 20 \] To solve for \( x \), follow these steps: 1. Subtract 11 from both sides of the equation: \[ e^{4 - 7x} = 9 \] 2. Take the natural logarithm (ln) of both sides: \[ \ln(e^{4 - 7x}) = \ln(9) \] 3. Because the natural logarithm and the exponential function are inverse operations, \( \ln(e^{y}) = y \). Hence: \[ 4 - 7x = \ln(9) \] 4. To isolate \( x \), subtract 4 from both sides: \[ -7x = \ln(9) - 4 \] 5. Finally, divide both sides by -7 to solve for \( x \): \[ x = \frac{\ln(9) - 4}{-7} \] So, \[ x = \frac{4 - \ln(9)}{7} \] Now you can use a calculator to find the approximate value of \( x \).
Solving an Exponential Equation
This is an equation involving an exponential function. To solve the equation \(e^{4 - 7x} + 11 = 20\), we need to isolate the term with the exponent and then solve for \(x\). Here are the steps to solve the equation: 1. Subtract 11 from both sides to isolate the exponential term: \(e^{4 - 7x} = 9\) 2. Take the natural logarithm (ln) of both sides to remove the base \(e\) and get the exponent by itself: \( \ln(e^{4 - 7x}) = \ln(9) \) 3. The natural logarithm and the exponential function are inverse functions, so: \(4 - 7x = \ln(9)\) 4. Solve for \(x\): \(7x = 4 - \ln(9)\) \(x = \dfrac{4 - \ln(9)}{7}\) 5. To find the numeric value, we can use a calculator: The natural logarithm of 9 (to two decimal places) is approximately 2.20. \(x \approx \dfrac{4 - 2.20}{7}\) \(x \approx \dfrac{1.80}{7}\) \(x \approx 0.2571\) So the approximate solution to the original equation is \(x \approx 0.2571\).
Solving Exponential Equation
The image shows the following exponential equation: e^(4 - 7x) + 11 = 20 To solve the equation for x, we'll follow these steps: 1. First, isolate the exponential term: e^(4 - 7x) = 20 - 11 e^(4 - 7x) = 9 2. Next, we would take the natural logarithm (ln) of both sides to get the exponent by itself: ln(e^(4 - 7x)) = ln(9) 3. Since ln(e^y) = y for any y (because ln and e are inverse functions), we can simplify the left-hand side: 4 - 7x = ln(9) 4. Lastly, we'll solve for x: 7x = 4 - ln(9) x = (4 - ln(9)) / 7 To find the numerical value, compute: ln(9) ≈ 2.1972 So, x ≈ (4 - 2.1972) / 7 x ≈ 1.8028 / 7 x ≈ 0.2575 Therefore, the solution to the equation is x ≈ 0.2575.
Solving Exponential Equation with Natural Logarithm
The equation in the image is: \( e^{4 - 7x} + 11 = 20 \) To solve for \( x \), follow these steps: 1. Isolate the exponential term by subtracting 11 from both sides of the equation: \( e^{4 - 7x} = 9 \) 2. Take the natural logarithm of both sides to eliminate the base \( e \): \( \ln( e^{4 - 7x} ) = \ln(9) \) 3. Use the logarithmic property \( \ln(e^y) = y \) to simplify the left side: \( 4 - 7x = \ln(9) \) 4. Isolate \( x \) by subtracting 4 from both sides and then dividing by -7: \( - 7x = \ln(9) - 4 \) \( x = \frac{\ln(9) - 4}{-7} \) 5. Calculate the value of \( x \): \( x \approx \frac{2.1972 - 4}{-7} \) \( x \approx \frac{-1.8028}{-7} \) \( x \approx 0.2575 \) Therefore, the solution for \( x \) is approximately 0.2575.
Solving an Exponential Equation with Two Variables
The image shows an exponential equation: e^(t + 7x) + 11 = 20. To solve for the variables t and x in this equation, follow these steps: 1. Isolate the exponential term on one side of the equation: e^(t + 7x) = 20 - 11 e^(t + 7x) = 9 2. Since e^(t + 7x) = 9, take the natural logarithm of both sides to get rid of the exponential base 'e': ln(e^(t + 7x)) = ln(9) 3. By the property of logarithms, ln(e^y) = y for any y, we have: t + 7x = ln(9) From this point, the equation cannot be solved for a unique solution because we have a single equation with two variables. To find specific values for t and x, we would need a second independent equation. If there is additional information or another equation provided, you can use it to solve for the unknowns. Otherwise, you have an infinite set of solutions depending on the value of one of the variables.
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