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Chapter 3 (Course Work Question 57, 3.4.41 HW Scorre \( 88.18 \%, 56.43 \) of 64 points Pat 1 of 2 Points: 0.5 of 1 Solve the following exponential equation. Express the solution in terms of natural logarithms or common logarithms. Then, use a calculator to obtain a decimal approximation for the solution. \[ 2^{5 x+6}=5^{x-2} \] The solution set expressed in terms of logarithms is \( \square \) \}. (Use a comma to separate answers as needed. Simplify your answer. Use integers or fractions for any numbers in the expression. Use In for natural logarithm and log for common loganithm.)

Ask by Guzman Lane. in Saudi Arabia
Mar 09,2025

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The solution in terms of natural logarithms is \( x = \frac{-2\ln 5 - 6\ln 2}{5\ln 2 - \ln 5} \), and its decimal approximation is approximately \( x \approx -3.97 \).

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To solve the exponential equation \(2^{5x+6} = 5^{x-2}\), we can take the logarithm of both sides. Let's use natural logarithms for this example: \[ \ln(2^{5x+6}) = \ln(5^{x-2}) \] Using the property of logarithms that states \(\ln(a^b) = b\ln(a)\), we rewrite the equation as: \[ (5x + 6) \ln(2) = (x - 2) \ln(5) \] Next, we can expand and rearrange the equation: \[ 5x \ln(2) + 6 \ln(2) = x \ln(5) - 2 \ln(5) \] Bringing all terms involving \(x\) to one side gives us: \[ 5x \ln(2) - x \ln(5) = -2 \ln(5) - 6 \ln(2) \] Factoring out \(x\): \[ x(5 \ln(2) - \ln(5)) = -2 \ln(5) - 6 \ln(2) \] Now we can isolate \(x\): \[ x = \frac{-2 \ln(5) - 6 \ln(2)}{5 \ln(2) - \ln(5)} \] That's our solution in terms of natural logarithms. For a decimal approximation, substitute values for \(\ln(2)\) and \(\ln(5)\): \[ \ln(2) \approx 0.693147, \quad \ln(5) \approx 1.609438 \] Calculating gives: \[ x \approx \frac{-2(1.609438) - 6(0.693147)}{5(0.693147) - 1.609438} \] Now, plug in the values and compute the approximation using a calculator. Your solution set expressed in terms of logarithms is: \[ x = \frac{-2 \ln(5) - 6 \ln(2)}{5 \ln(2) - \ln(5)} \] Now, let's do a fun fact about logarithms! Did you know that the logarithm was invented by John Napier in the early 17th century? He developed it as a way to simplify computations, particularly multiplication and division, which were very tedious before calculators and computers. Logarithms turned multiplications into additions, making calculations far more manageable! And as for real-world applications, logarithms are everywhere! They're used in everything from measuring the intensity of earthquakes on the Richter scale to pH in chemistry and even in calculating compound interest in finance! So, the next time you hear about logarithms, just remember how crucial they are in various fields of study and real-life applications.

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