Example Question - solving logarithmic equations
Here are examples of questions we've helped users solve.
Solving Logarithmic Equations with Quadratic Terms
To solve the logarithmic equation \(\log_2{(x^2 - 4)} + 1 = \log_2{(2 - x)}\), you can start by simplifying the equation using properties of logarithms. The equation currently has logarithms with the same base on both sides. To isolate the logarithmic expressions, you can start by dealing with the "+1" on the left side. Since it's not part of the logarithmic term, you can consider using the property that adding 1 to a log is equivalent to multiplying its argument by the base (in this case, 2). The equation can be rewritten as: \[\log_2{(x^2 - 4)} = \log_2{(2 - x)} - 1\] Converting the subtraction of 1 on the right side to the equivalent form in logarithmic expression, we get: \[\log_2{(x^2 - 4)} = \log_2{\frac{2 - x}{2}}\] Now that we have a log on each side with the same base and there are no additional constants, we can equate the arguments (since if \(\log_b{A} = \log_b{B}\), then \(A = B\)): \[x^2 - 4 = \frac{2 - x}{2}\] Now let's solve for \(x\). Multiply both sides by 2 to get rid of the fraction: \[2(x^2 - 4) = 2 - x\] \[2x^2 - 8 = 2 - x\] Now bring all terms to one side to form a quadratic equation: \[2x^2 + x - 10 = 0\] To solve this quadratic equation, you can use the quadratic formula, where \(a = 2\), \(b = 1\), and \(c = -10\): \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Plug in the values: \[x = \frac{-1 \pm \sqrt{1^2 - 4(2)(-10)}}{2(2)}\] \[x = \frac{-1 \pm \sqrt{1 + 80}}{4}\] \[x = \frac{-1 \pm \sqrt{81}}{4}\] \[x = \frac{-1 \pm 9}{4}\] This gives us two possible solutions: \[x = \frac{-1 + 9}{4} = \frac{8}{4} = 2\] \[x = \frac{-1 - 9}{4} = \frac{-10}{4} = -2.5\] However, we need to check these solutions to ensure they are valid in the original logarithmic equation (we must avoid negative numbers or zero inside the logarithms, as the log of a non-positive number is undefined). The solution \(x = 2\) would make the expression inside the first logarithm zero (\(x^2 - 4 = 2^2 - 4 = 4 - 4 = 0\)), which is not allowed. Therefore, \(x = 2\) is not a valid solution. The solution \(x = -2.5\) would make the expression inside the first logarithm positive (\((-2.5)^2 - 4 = 6.25 - 4 = 2.25\)), and the expression inside the second logarithm negative (\(2 - (-2.5) = 4.5\)), which is also allowed. Therefore, \(x = -2.5\) is the valid solution for the original equation.
Solving Logarithmic Equations by Converting to Exponential Form
To solve the given logarithmic equation \(\log_4(5x + 9) = 3\), you can begin by converting the logarithmic form to exponential form. The basic logarithm rule states that if \(\log_b(A) = C\), then \(b^C = A\). Applying this rule here gives: \(4^3 = 5x + 9\) Now, calculate the left side: \(4^3 = 4 \times 4 \times 4 = 64\) So the equation becomes: \(64 = 5x + 9\) Now, subtract 9 from both sides to isolate the term containing \(x\): \(64 - 9 = 5x\) \(55 = 5x\) Finally, divide both sides by 5 to solve for \(x\): \(x = 55 / 5\) \(x = 11\) Therefore, the solution to the equation is \(x = 11\).
Solving Logarithmic Equations by Converting to Exponential Form
To solve the logarithmic equation \(\log(2x + 4) = 2\), we can rewrite the equation in its exponential form. The base of the logarithm is 10 by default when no base is specified. With this in mind, the equation becomes: \(10^{\log(2x + 4)} = 10^2\) Since \(10^{\log(x)} = x\), we have: \(2x + 4 = 10^2\) \(2x + 4 = 100\) Next, we solve for \(x\) by isolating the variable: \(2x = 100 - 4\) \(2x = 96\) Divide both sides by 2 to find \(x\): \(x = \frac{96}{2}\) \(x = 48\) Hence, the solution to the equation \(\log(2x + 4) = 2\) is \(x = 48\).
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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