Example Question - 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>
Logarithm Equivalence Proof
<p>Given \(\log\frac{81}{8} = 2\log\frac{2}{3} + 3\log\frac{1}{3}\)</p> <p>Apply logarithm properties: \(\log a^b = b \cdot \log a\)</p> <p>\(\log\frac{3^4}{2^3} = \log\left(\frac{2}{3}\right)^2 + \log\left(\frac{1}{3}\right)^3\)</p> <p>\(\log\frac{3^4}{2^3} = \log\left(\frac{2^2}{3^2}\right) + \log\left(\frac{1}{3^3}\right)\)</p> <p>\(\log\frac{3^4}{2^3} = \log\frac{2^2}{3^2} + \log\frac{1}{3^3}\)</p> <p>\(\log\frac{3^4}{2^3} = \log\frac{2^2}{3^2} + \log\frac{1}{3^3}\)</p> <p>Combine logs on the right using \(\log a + \log b = \log(ab)\):</p> <p>\(\log\frac{3^4}{2^3} = \log\left(\frac{2^2}{3^2} \cdot \frac{1}{3^3}\right)\)</p> <p>\(\log\frac{3^4}{2^3} = \log\frac{2^2}{3^5}\)</p> <p>Simplify the right side:</p> <p>\(\log\frac{3^4}{2^3} = \log\frac{3^4}{2^3}\)</p> <p>Since the logs with the same base are equal, their arguments must be equal, proving the statement:</p> <p>\(\frac{3^4}{2^3} = \frac{3^4}{2^3}\)</p> <p>Hence, the given equation is proved to be correct.</p>
Domain and Range Determination for Given Functions
<p>Untuk soal (a):</p> <p>D_{f(x,y)} = \{(x, y) \in \mathbb{R}^2 | x^2 - y^2 - 9 \geq 0\}</p> <p>R_{f(x,y)} = \{f(x,y) \in \mathbb{R} | f(x,y) = 4-x^2-y^2 + \frac{1}{2}\log(x^2 - y^2 - 9), (x, y) \in D_{f(x,y)} \}</p> <p>Untuk soal (b):</p> <p>D_{f(x,y)} = \{(x, y) \in \mathbb{R}^2 | x^2 + y^2 - 1 \neq 0\}</p> <p>R_{f(x,y)} = \{f(x,y) \in \mathbb{R} | f(x,y) = \frac{\sqrt{x^2 - y^2}}{x^2 + y^2 - 1}, (x, y) \in D_{f(x,y)} \}</p>
Logarithmic Equation Problem
\[ \text{原方程式為：} \log_7(7^{2y} - 18) = y + 1 \] \[ \text{假設} \log_7(7^{2y} - 18) = a \text{，則} 7^a = 7^{2y} - 18 \] \[ \text{原方程式可寫作：} 7^{y + 1} = 7^{2y} - 18 \] \[ \text{注意到} 7^{y+1} = 7^y \cdot 7 \text{，帶回方程式中：} \] \[ 7^y \cdot 7 = 7^{2y} - 18 \] \[ \text{令} x = 7^y \text{，則上式轉變為：} \] \[ 7x = x^2 - 18 \] \[ \text{此為一元二次方程，整理得：} \] \[ x^2 - 7x - 18 = 0 \] \[ \text{解此一元二次方程得：} \] \[ (x - 9)(x + 2) = 0 \] \[ x = 9 \text{ 或 } x = -2 \] \[ \text{因為} 7^y \text{總是正數，所以} x \neq -2 \] \[ \text{所以} x = 9 \text{即} 7^y = 9 \text{，將其轉為對數形式得：} \] \[ y = \log_7 9 \] \[ \text{所以答案是 C. } \log_7 9 \]
Logarithmic Equation Simplification
<p>给定等式为：</p> <p>\[\frac{4\log a + 7\log\sqrt[3]{a}}{4\log a - 7\log\sqrt[3]{a}}\]</p> <p>首先，我们简化对数：</p> <p>\[\log\sqrt[3]{a} = \log a^{\frac{1}{3}} = \frac{1}{3}\log a\]</p> <p>带入简化的对数：</p> <p>\[\frac{4\log a + 7\cdot\frac{1}{3}\log a}{4\log a - 7\cdot\frac{1}{3}\log a} = \frac{4\log a + \frac{7}{3}\log a}{4\log a - \frac{7}{3}\log a}\]</p> <p>合并对数项：</p> <p>\[\frac{\frac{19}{3}\log a}{\frac{5}{3}\log a}\]</p> <p>由于分子和分母都有对数项，我们可以约去这些项：</p> <p>\[\frac{19}{5}\]</p> <p>因此，最终答案为 B. \[\frac{19}{5}\]</p>
Calculating 4 to the power of 3-3b
题目提供了以下的信息：\(2^5=5\), \( \log_8 3 = \frac{1}{6} \)，要我们计算 \(4^{3-3b}\)。 首先，我们需要将4的幂表示为2的幂，因为我们知道与2相关的信息。由于 \(4 = 2^2\)，我们可以重写 \(4^{3-3b}\) 为 \( (2^2)^{3-3b} = 2^{2(3-3b)} = 2^{6-6b} \)。 我们已经知道 \(2^5 = 5\)，但我们需要计算的是 \(2^{6-6b}\)。现在我们需要一个与b相关的表达式，我们可以利用 \( \log_8 3 = \frac{1}{6} \) 来得到b的值。 根据对数的定义，我们可以将 \( \log_8 3 = \frac{1}{6} \) 重写为 \( 8^\frac{1}{6} = 3 \)，而 \(8 = 2^3\)，所以 \( (2^3)^\frac{1}{6} = 3 \)，或者 \(2^\frac{1}{2} = 3\)。 这意味着 \(2^b = 3 \rightarrow b = \frac{1}{2}\)。现在我们已经知道了b的值，我们可以将其代入 \(2^{6-6b}\)： \(2^{6-6b} = 2^{6-6(\frac{1}{2})} = 2^{6-3} = 2^3 \)。 因为我们没有 \(2^3\) 直接的值，所以我们无法根据已知信息计算出具体的数字。如果题目中只提供了 \(2^5 = 5\) 和 \( \log_8 3 = \frac{1}{6} \)，我们没有足够的信息来确定 \(2^3\) 的实际数值，需要额外的信息才能解决这个问题。
Inverse of Exponential Function
The question asks for the inverse of the exponential function \( y = 4^x \). To find the inverse function, we typically swap \( x \) and \( y \) and then solve for \( y \). Starting with \( y = 4^x \), we swap \( x \) and \( y \) to get \( x = 4^y \). Now we need to solve for \( y \). To do this, we take the logarithm base 4 of both sides: \( \log_4(x) = \log_4(4^y) \). Using the property of logarithms that \( \log_b(b^a) = a \), we simplify the right side to get: \( \log_4(x) = y \). So, the inverse function of \( y = 4^x \) is \( y = \log_4(x) \). The correct answer is: C. \( y = \log_4(x) \).
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