Example Question - logarithms
Here are examples of questions we've helped users solve.
Complex Logarithmic Expression Simplification
<p>\[\text{Given expression: } \frac{\log 343}{1 + \frac{1}{2}\log(49) - \frac{1}{3}\log\left(\frac{1}{125}\right)}\]</p> <p>\[= \frac{\log 7^3}{1 + \frac{1}{2}\log(7^2) - \frac{1}{3}\log\left(5^{-3}\right)}\]</p> <p>\[= \frac{3\log 7}{1 + \frac{1}{2}\cdot 2 \log 7 - \frac{1}{3} \cdot (-3) \log 5}\]</p> <p>\[= \frac{3\log 7}{1 + \log 7 + \log 5}\]</p> <p>\[= \frac{3\log 7}{\log 7 + \log 5 + \log 10}\]</p> <p>\[= \frac{3\log 7}{\log(7 \cdot 5 \cdot 10)}\]</p> <p>\[= \frac{3\log 7}{\log 350}\]</p> <p>\[\text{Since } \log_a b = \frac{1}{\log_b a}, \text{ swap the logs:}\]</p> <p>\[= \frac{3}{\log_{7} 350}\]</p> <p>\[= \frac{3}{\log_{7} 7^2 + \log_{7} 5}\]</p> <p>\[= \frac{3}{2 + \log_{7} 5}\]</p> <p>\[\text{Without information on the value of } \log_{7} 5, \text{ this is the simplest form.}\]</p>
Logarithmic Equation Problem
<p>\text{已知等式：}\log_3(x + 4) + \log_3(x - 2) = 3.</p> <p>\text{应用对数法则合并同底对数：}\log_3[(x + 4)(x - 2)] = 3.</p> <p>\text{化简等式得：}\log_3(x^2 + 2x - 8) = 3.</p> <p>\text{消去对数得：}x^2 + 2x - 8 = 3^3.</p> <p>\text{解方程：}x^2 + 2x - 8 = 27.</p> <p>x^2 + 2x - 35 = 0.</p> <p>(x + 7)(x - 5) = 0.</p> <p>\text{所以解为：}x = -7 \text{ 或 } x = 5.</p>
Solving Exponential Equation with Logarithms
The equation provided in the image is: 5^(3x) + 2 = 8 To solve for x, you should follow these steps: 1. Subtract 2 from both sides to isolate the exponential term: 5^(3x) + 2 - 2 = 8 - 2 5^(3x) = 6 2. Take the logarithm of both sides of the equation. It's often easier to use the logarithm with the same base as the exponent, but you can use the natural logarithm (ln) or common logarithm (log base 10): ln(5^(3x)) = ln(6) 3. Use the power rule of logarithms, which states that ln(a^b) = b*ln(a): 3x * ln(5) = ln(6) 4. Divide both sides by 3*ln(5) to solve for x: x = ln(6) / (3 * ln(5)) Now, we can compute the value of x numerically. Note that since you're asked to round to the nearest thousandth, you should use a calculator to ensure accuracy. x ≈ ln(6) / (3 * ln(5)) Using a calculator: x ≈ ln(6) / (3 * ln(5)) x ≈ 1.791759469 / (3 * 1.609437912) x ≈ 1.791759469 / 4.828313737 x ≈ 0.371072 Rounding to the nearest thousandth: x ≈ 0.371 So, x is approximately 0.371 when rounded to the nearest thousandth.
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
Solving Logarithmic Expressions with Log Rules
To solve the expression given in the image, \( \log_{c^2}(\frac{a^5b}{c^2}) \), we can apply log rules (quotient rule, power rule, and change of base formula). The quotient rule states that \( \log(\frac{x}{y}) = \log(x) - \log(y) \). The power rule states that \( \log(x^a) = a \log(x) \). The change of base formula states that \( \log_b(a) = \frac{\log_c(a)}{\log_c(b)} \). Let's break down the expression by using these rules: 1. Apply the quotient rule to the argument of the log. \( \log_{c^2}(a^5b) - \log_{c^2}(c^2) \) 2. Apply the power rule to \( a^5 \) and \( c^2 \) within the log expressions. \( 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 \log_{c^2}(c) \) 3. Notice that \( \log_{c^2}(c^2) = 1 \) because the log base and the argument are the same value raised to the same power. Hence, the expression becomes: \( 5 \log_{c^2}(a) + \log_{c^2}(b) - 2 \) 4. Apply the change of base formula to the two remaining log terms. Since we want to express everything in terms of the base c, we get: \( 5 \frac{\log_{c}(a)}{\log_{c}(c^2)} + \frac{\log_{c}(b)}{\log_{c}(c^2)} - 2 \) 5. Simplify by recognizing that \( \log_{c}(c^2) = 2 \). \( 5 \frac{\log_{c}(a)}{2} + \frac{\log_{c}(b)}{2} - 2 \) 6. Simplify the fractions. \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \) 7. Looking at the answer choices given, we can see that our derived expression matches choice D when we distribute the 2 outside of the log: \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \cdot 1 \) \( \frac{5}{2} \log_{c}(a) + \frac{1}{2} \log_{c}(b) - 2 \log_{c}(c) \) Since \( \log_{c}(c) = 1 \), the 2 just becomes \( 2 \log_{c}(c) \), and thus the answer is: \( \frac{5}{2} \cdot \log_{c}(a) + \frac{1}{2} \cdot \log_{c}(b) - 2 \log_{c}(c) \) So, the correct answer is D: \( 5 \log_{c}(a) + \log_{c}(b) - 2 \log_{c}(c) \)
Converting Exponential to Logarithmic Form
The image shows a handwritten question asking to convert the form of the following equation, specifically from exponential form to logarithmic form, and vice versa. The given equation is: 7^x = 2401 To convert from exponential form to logarithmic form, you apply the basic definition of logarithms. If a^b = c, then log_a(c) = b. Applying this to the given equation, you get: log_7(2401) = x. To find the value of x, you need to determine what power of 7 gives you 2401. 2401 is 7 raised to the fourth power since 7 * 7 * 7 * 7 = 2401. Therefore, the logarithmic form of the equation is: log_7(2401) = 4. Hence, x = 4.
Solving Logarithmic Equation for x
This question is an equation involving logarithms that we need to solve for x: \( \frac{\log(32)}{\log(x)} + \frac{\log(16)}{\log(x)} = \log(x) \) Firstly, we can combine the terms on the left-hand side using the fact that \( \log(a) + \log(b) = \log(a \cdot b) \). However, since these logarithms have the same denominator, we can combine them as follows: \( \frac{\log(32) + \log(16)}{\log(x)} = \log(x) \) Next, we use the property of logarithms that states \( \log(a) + \log(b) = \log(ab) \) to combine the numerators: \( \frac{\log(32 \cdot 16)}{\log(x)} = \log(x) \) \( \frac{\log(512)}{\log(x)} = \log(x) \) Now we have a single logarithm in the numerator. We can simplify this further by using the property \( \log_b(a) = c \) implies \( b^c = a \): Let's equate the left side to the right side. This gives us an equation of the form: \( \log(512) = \log(x^2) \) Since the logarithms are in the same base and the equation equates one log to another, we can deduce that the arguments must be equal: \( 512 = x^2 \) Taking the square root of both sides gives us: \( x = \sqrt{512} \) \( x = 16 \times \sqrt{2} \) \( x = 16 \times 1.414... \) (since \( \sqrt{2} \) is approximately 1.414) \( x \approx 22.63 \) Therefore, the value of \( x \approx 22.63 \). Please note this value has been rounded to two decimal places.
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