Example Question - logarithmic properties
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Simplification of a Logarithmic Expression
<p>\(\text{Diberikan } \frac{\log_2{20} - \log_2{5}}{\log_4}\)</p> <p>\(\text{Gunakan sifat logaritma untuk menyederhanakan }\log_2{20} - \log_2{5} \text{ menjadi } \log_2{\frac{20}{5}}\)</p> <p>\(\log_2{\frac{20}{5}} = \log_2{4}\)</p> <p>\(\text{Kemudian, ubah basis logaritma }\log_4 \text{ menggunakan rumus perubahan basis: } \log_4 = \frac{\log_2{4}}{\log_2{4}}\)</p> <p>\(\log_4 = \frac{2}{\log_2{4}}\)</p> <p>\(\text{Jadi, ekspresi keseluruhannya menjadi } \frac{\log_2{4}}{\frac{2}{\log_2{4}}}\)</p> <p>\(\text{Sederhanakan untuk mendapatkan hasil: } \frac{(\log_2{4})(\log_2{4})}{2}\)</p> <p>\(\text{Diketahui bahwa } \log_2{4} = 2 \text{ sehingga menjadi }\)</p> <p>\(\frac{2 \cdot 2}{2} = 2\)</p>
Solving an Exponential Expression
<p>Данное выражение: \( 4^{log_3{2}} \)</p> <p>По определению логарифма \( a^{log_a{b}} = b \), поэтому:</p> <p>\( 4^{log_3{2}} = 2^{log_3{4}} \)</p> <p>Используем свойство логарифмов \( log_a{b^n} = n \cdot log_a{b} \) и выразим \( log_3{4} \) как \( 2 \cdot log_3{2} \):</p> <p>\( 2^{log_3{4}} = 2^{2 \cdot log_3{2}} \)</p> <p>Тогда используем свойство экспонент: \( a^{m \cdot n} = (a^m)^n \):</p> <p>\( 2^{2 \cdot log_3{2}} = (2^{log_3{2}})^2 \)</p> <p>По определению логарифма возводим обратно:</p> <p>\( (3^{log_3{2}})^2 = 2^2 \)</p> <p>Таким образом, итоговый ответ:</p> <p>\( 2^2 = 4 \)</p>
Solving a Logarithmic Equation with Logarithmic Properties
To solve the given logarithmic equation: \[\log_{2}(x^2 - 4) + 1 = \log_{2}(2^{-x})\] Let's first simplify the right-hand side of the equation. We know from the laws of logarithms that if \(a = \log_b{(c)}\), then \(b^a = c\). Applying this to our case: \[\log_{2}(2^{-x}) = -x\] This is because \(2^{-x}\) is the value that \(2\) needs to be raised to in order to equal \(2^{-x}\), which clearly is \(-x\). Now, let us use the fact that adding \(1\) to a \(\log_2\) value is the same as multiplying its argument by \(2\): \[\log_{2}(x^2 - 4) + 1 = \log_{2}((x^2 - 4) \cdot 2)\] With this, the equation becomes: \[\log_{2}(2x^2 - 8) = \log_{2}(2^{-x})\] Since the logs have the same base, their arguments must be equal for the equation to hold true: \[2x^2 - 8 = 2^{-x}\] Now solve this equation for \(x\). Depending on the complexity, this might require numerical methods since it doesn't simplify to an easily solvable polynomial. However, let's try to see if we can find any solutions by inspection or simplification. Rearranging the terms gives us: \[2x^2 - 2^{-x} - 8 = 0\] Unfortunately, this equation involves both a quadratic term and an exponential term with a negative exponent, which makes it a transcendental equation without an algebraic solution. To solve this, you would usually use numeric methods like Newton's method or a graphing calculator to approximate the solution. Since solving this equation analytically is not simple, let's use numerical methods. To preserve the accuracy and process of complex calculation that cannot be performed as text, we might usually suggest using computational tools such as a graphing calculator or software like Wolfram Alpha, which can handle equations of this form and provide numerical solutions.
Solving and Combining Logarithmic Expressions
To solve the expression given, 2 log 5 + 5 log x, and express it as a single logarithm, we will utilize the properties of logarithms: 1. The power rule: log(a^b) = b * log(a) 2. The product rule: log(a) + log(b) = log(a * b) Let's apply these rules step by step: The first term 2 log 5 can be rewritten using the power rule as: log(5^2) = log(25) The second term 5 log x can be rewritten using the power rule as: log(x^5) Now, adding the two log terms using the product rule: log(25) + log(x^5) = log(25 * x^5) Now the expression is written as a single logarithm: log(25 * x^5) Looking at the answer choices, the correct answer is B. log(25 * x^5).
Simplifying Logarithmic Expressions
The expression provided in the image, \(\log\left(\frac{a^3b}{c^2}\right)\), can be simplified using the properties of logarithms: The Quotient Rule: \(\log(x/y) = \log(x) - \log(y)\) The Power Rule: \(\log(x^k) = k \cdot \log(x)\) So let's break down the expression step by step: \(\log\left(\frac{a^3b}{c^2}\right)\) = \( \log(a^3b) - \log(c^2) \) (using the Quotient Rule) = \( \log(a^3) + \log(b) - \log(c^2) \) (using the Product Rule: \(\log(xy) = \log(x) + \log(y)\)) = \( 3\cdot\log(a) + \log(b) - 2\cdot\log(c) \) (using the Power Rule) Thus, the correct answer to the expression given in the image is: \(\log(a) + 3\log(b) - 2\log(c)\) This corresponds to answer choice D in the image.
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.
Analyzing Exponential and Logarithmic Functions
Trước tiên, để giải quyết các câu hỏi từ hình ảnh, chúng ta sẽ xem xét từng câu một. Câu 1: Với \( a \) là số thực dương tuỳ ý, biểu thức \( a^{a^a} \) là Ta biết rằng biểu thức \( a^{a^a} \) có thể viết lại như sau: \( a^{(a^a)} \), theo quy tắc lũy thừa, giá trị của số mũ được xác định bởi biểu thức trong ngoặc trước tiên. Điều này không giống với bất kỳ đáp án nào đã cho, vì các đáp án đều có dạng \( a^{\text{số mũ}} \) với số mũ không có ngoặc. Có thể hình ảnh này chưa hiển thị đầy đủ các phương án lựa chọn hoặc câu hỏi có thể có lỗi. Tuy nhiên, dựa vào các phương án lựa chọn được cung cấp, đây có vẻ như là một câu hỏi về các đặc tính của lũy thừa. Nếu chúng ta giả sử đề bài cần chúng ta viết lại biểu thức \( a^{a^a} \) dưới dạng một lũy thừa đơn giản hơn, vẫn không có đáp án nào đúng với cách biểu diễn này. Câu 2: Với \( a \) là số thực dương tuỳ ý, \( \log_a {a^2} \) bằng Để giải bài toán này, ta sử dụng định nghĩa cơ bản của logarit: \( \log_b {x} \) là số \( y \) sao cho \( b^y = x \). Áp dụng định nghĩa này vào đề bài, ta có: \( \log_a {a^2} \) là số mũ cần thiết để \( a \) được nâng lên quyền lực để nhận được \( a^2 \). Vì vậy, ta tìm số mũ \( y \) sao cho: \( a^y = a^2 \) Từ đây có thể thấy \( y = 2 \). Vậy \( \log_a {a^2} = 2 \). Chọn đáp án là: A. \( 2 \log_a {a} \) B. \( 2 + \log_a {a} \) C. \( \frac{1}{2} \log_a {a} \) D. \( -\frac{1}{2} \log_a {a} \) Đáp án đúng là B. \( 2 + \log_a {a} \). Tuy nhiên, đáp án này là không chính xác theo cách chúng ta vừa giải, vì \( 2 + \log_a {a} \) thực sự sẽ bằng 2 (do \( \log_a {a} \) luôn bằng 1). Vì thế, có vẻ như lựa chọn đúng phải là \( 2 \) đơn giản, không kèm theo \( \log_a {a} \), có nghĩa là đề có thể đã bị in sai hoặc ta hiểu sai ý của đề. Câu 3: Tập xác định của hàm số \( y = \log_2 {3x} \) là Hàm logarit chỉ xác định khi giá trị bên trong logarit là một số dương. Vì vậy ta cần \( 3x > 0 \), điều này có nghĩa là \( x > 0 \). Vậy tập xác định của hàm số là (0, +∞), đáp án: A. \( (-\infty, 0) \) B. \( (0, +\infty) \) C. \( 0 \) D. \( [1, +∞) \) Đáp án đúng là B. \( (0, +∞) \).
Solving Logarithmic Expressions with Properties
To solve the expression given in the image as a single logarithm, we can use the properties of logarithms: The original expression given is: \[ \frac{1}{4} \ln 4 - 2 \ln x \] Firstly, we can use the power rule of logarithms which states that \( \log_b (a^n) = n \log_b a \). We can apply this rule on the first term to bring down the 1/4 as a power of 4 inside the logarithm: \[ = \ln 4^{1/4} - 2 \ln x \] \[ = \ln 2 - 2 \ln x \] (Note: \( 4^{1/4} \) is the fourth root of 4, which is equal to 2.) Next, we use another property of logarithms: the coefficient in front of a logarithm can be converted into a power inside the logarithm, which is: \[ \log_b a^c = c \log_b a \] Applying this to the second term: \[ = \ln 2 - \ln x^2 \] Now, we use the subtraction rule for logarithms, which states: \( \log_b a - \log_b c = \log_b \frac{a}{c} \). We'll apply this to the expression: \[ = \ln \frac{2}{x^2} \] Looking at the options provided in the image, the one that matches is B: \[ \ln \left( \frac{2}{x^2} \right) \]
Solving Logarithmic Equation
The equation shown in the image is: \[ \frac{\log_2 32 + \log_2 16}{\log_2 x} = \log_2 x \] To solve for \( x \), we can use logarithmic properties to simplify the left side of the equation and then isolate \( x \). Here are the steps: 1. Simplify the logarithms on the left side of the equation: \[ \log_2 32 = 5 \] (because \( 2^5 = 32 \)) \[ \log_2 16 = 4 \] (because \( 2^4 = 16 \)) So the equation becomes: \[ \frac{5 + 4}{\log_2 x} = \log_2 x \] 2. Further simplify the numerator: \[ \frac{9}{\log_2 x} = \log_2 x \] 3. To isolate \( x \), we can equate the exponents by removing the logarithms. Multiply both sides by \( \log_2 x \) to get rid of the denominator: \[ 9 = \log_2 x \cdot \log_2 x \] \[ 9 = (\log_2 x)^2 \] 4. Take the square root of both sides: \[ \sqrt{9} = \sqrt{(\log_2 x)^2} \] \[ 3 = |\log_2 x| \] This gives us two possibilities, because the absolute value of \( \log_2 x \) can be either \( 3 \) or \( -3 \). Case 1: \( \log_2 x = 3 \) \[ 2^3 = x \] \[ x = 8 \] Case 2: \( \log_2 x = -3 \) \[ 2^{-3} = x \] \[ x = \frac{1}{8} \] However, in the context of the original equation, a negative logarithm would not make sense because it would imply that \( x \) is between 0 and 1, but such \( x \) values would result in a negative denominator in the original equation, which cannot equal a positive \( \log_2 x \) as seen on the right side. Therefore, we discard the second case \( x = \frac{1}{8} \). The solution to the original equation is \( x = 8 \).
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: \[ x = \sqrt{48a} \]
Solving a Logarithmic Expression Step by Step
The image shows a mathematical expression involving logarithms. We will solve the expression step by step: The expression is: \[ \frac{1}{2} \log_2 \left( \frac{4}{3} \right) - 2\log_2 \left( \frac{9}{4} \right) \] First, let's simplify the logarithmic terms using the properties of logarithms: 1) Use the power rule of logarithms, which states that \(\log_b(m^n) = n\cdot\log_b(m)\), to simplify the second term: \[ 2\log_2 \left( \frac{9}{4} \right) = \log_2 \left( \left(\frac{9}{4}\right)^2 \right) \] 2) Calculate the square of \(\frac{9}{4}\): \[ \left(\frac{9}{4}\right)^2 = \frac{81}{16} \] Now, we can rewrite the expression as: \[ \frac{1}{2} \log_2 \left( \frac{4}{3} \right) - \log_2 \left( \frac{81}{16} \right) \] 3) Apply the quotient rule for logarithms, which states that \(\log_b \left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\), to both logarithmic terms: \[ \frac{1}{2} \left( \log_2(4) - \log_2(3) \right) - \left( \log_2(81) - \log_2(16) \right) \] 4) Simplify the logarithms using the fact that \(\log_2(4) = 2\), \(\log_2(16) = 4\), and \(\log_2(81)\) is \(\log_2(3^4)\) which equals \(4\log_2(3)\): \[ \frac{1}{2} \left( 2 - \log_2(3) \right) - \left( 4\log_2(3) - 4 \right) \] 5) Simplify the expression by distributing and combining like terms: \[ 1 - \frac{1}{2}\log_2(3) - 4\log_2(3) + 4 \] \[ 5 - \left( \frac{1}{2}\log_2(3) + 4\log_2(3) \right) \] \[ 5 - \left( \frac{1}{2} + 4\right) \log_2(3) \] \[ 5 - \frac{9}{2} \log_2(3) \] Here is the simplified form of the original expression: \[ 5 - \frac{9}{2} \log_2(3) \] This is as simplified as the expression can get without knowing the numerical value of \(\log_2(3)\).
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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