Example Question - logarithm rules
Here are examples of questions we've helped users solve.
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.
Solving Logarithmic Equations with Multiple Steps
To solve the equation involving logarithms from the image provided, follow these steps: Given the equation: \[ \frac{\log_2{3a} + \log_2{16}}{\log_2{x}} = \log_2{x} \] 1. First apply the log rule \(\log_b{m} + \log_b{n} = \log_b{mn}\) to combine the logs in the numerator: \[ \frac{\log_2{(3a \cdot 16)}}{\log_2{x}} = \log_2{x} \] 2. Since \(16 = 2^4\), we can write the equation as: \[ \frac{\log_2{(3a \cdot 2^4)}}{\log_2{x}} = \log_2{x} \] \[ \frac{\log_2{48a}}{\log_2{x}} = \log_2{x} \] 3. Multiply both sides by \(\log_2{x}\) to get rid of the denominator: \[ \log_2{48a} = (\log_2{x})^2 \] 4. Recognize that \((\log_2{x})^2\) is the same as \(\log_2{x^2}\), and use the property that \(\log_b{m} = \log_b{n}\) implies \(m = n\): \[ 48a = x^2 \] 5. You can now solve for \(a\), given \(x\), or solve for \(x\) given \(a\). If you want to find a formula for \(a\) in terms of \(x\), divide by 48: \[ a = \frac{x^2}{48} \] Or to find \(x\) in terms of \(a\), take the square root of both sides: \[ x = \pm\sqrt{48a} \] In the context of logarithms, \(x\) is typically assumed to be positive, so the negative solution is often discarded, leaving: \[ x = \sqrt{48a} \]
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