Example Question - properties of logarithms
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Simplifying Logarithmic Expressions
<p>For the given logarithmic expression:</p> <p>\(\log 7 + \log 2 + \log \frac{1}{40} + \log \frac{1}{7}\)</p> <p>We can simplify using the properties of logarithms:</p> <p>Combine \(\log 7\) and \(\log \frac{1}{7}\) using \(\log a + \log \frac{1}{a} = \log 1 = 0\):</p> <p>\(\log 7 + \log \frac{1}{7} = \log 1 = 0\)</p> <p>Combine \(\log 2\) and \(\log \frac{1}{40}\) by rewriting \(\frac{1}{40}\) as \(2^{-3}\), then use the property \(\log a^b = b \log a\):</p> <p>\(\log 2 + \log 2^{-3} = \log 2 + (-3)\log 2 = \log 2 - 3 \log 2\)</p> <p>\(\log 2 - 3 \log 2 = -2 \log 2\)</p> <p>Since \(\log 7 + \log \frac{1}{7} = 0\), all that's left is:</p> <p>\(-2 \log 2\)</p>
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 Exponential Equations with Natural Logarithms
To solve the equation \( e^{4-7x} + 11 = 20 \), follow these steps: 1. Subtract 11 from both sides of the equation: \[ e^{4-7x} + 11 - 11 = 20 - 11 \] \[ e^{4-7x} = 9 \] 2. Now take the natural logarithm (ln) of both sides to solve for \( 4-7x \): \[ \ln(e^{4-7x}) = \ln(9) \] Using the property of logarithms, \( \ln(e^y) = y \), we have: \[ 4 - 7x = \ln(9) \] 3. Solve for \( x \): \[ -7x = \ln(9) - 4 \] \[ x = \frac{-\ln(9) + 4}{7} \] Use a calculator to compute the value of \( \ln(9) \), and then compute \( x \).
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