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Chapter 3 (Course Work Question 57, 3.4.41 HW Scoro: 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. \[ 5^{2 x+6}=2^{x-1} \] 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 logarithm.)

Ask by Jimenez Gardner. in Saudi Arabia
Mar 09,2025

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Answer

The solution set expressed in terms of logarithms is: \[ x = \frac{-\ln 2 - 6\ln 5}{2\ln 5 - \ln 2} \]

Solution

We start with the equation: \[ 5^{2x+6}=2^{x-1}. \] Taking the natural logarithm on both sides gives: \[ \ln\left(5^{2x+6}\right)=\ln\left(2^{x-1}\right). \] Using the logarithm power rule, \(\ln(a^b)=b\ln(a)\), we have: \[ (2x+6)\ln 5=(x-1)\ln 2. \] Expanding both sides: \[ 2x\ln 5+6\ln 5=x\ln 2-\ln 2. \] Now, collect the terms containing \(x\) on one side: \[ 2x\ln 5 - x\ln 2 = -\ln 2 - 6\ln 5. \] Factor out \(x\): \[ x (2\ln 5-\ln 2)=-\ln 2-6\ln 5. \] Solve for \(x\): \[ x=\frac{-\ln 2-6\ln 5}{2\ln 5-\ln 2}. \] For a decimal approximation, use the approximate values \(\ln 5 \approx 1.6094\) and \(\ln 2 \approx 0.6931\): \[ \text{Denominator: } 2\ln 5 - \ln 2 \approx 2(1.6094) - 0.6931 \approx 3.2188 - 0.6931 \approx 2.5257, \] \[ \text{Numerator: } -\ln 2 - 6\ln 5 \approx -0.6931 - 6(1.6094) \approx -0.6931 - 9.6564 \approx -10.3495. \] Thus, \[ x\approx \frac{-10.3495}{2.5257}\approx -4.098. \] The solution set expressed in terms of logarithms is: \[ \left\{x=\frac{-\ln 2-6\ln 5}{2\ln 5-\ln 2}\right\}. \]

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To solve the equation \(5^{2x+6}=2^{x-1}\), take the logarithm of both sides. You can use either the natural logarithm or common logarithm. Here, we'll use natural logarithms: \[ \ln(5^{2x+6}) = \ln(2^{x-1}) \] Using the power rule of logarithms \( \ln(a^b) = b\ln(a) \): \[ (2x + 6)\ln(5) = (x - 1)\ln(2) \] Now, distribute and rearrange to isolate \(x\): \[ 2x \ln(5) + 6 \ln(5) = x \ln(2) - \ln(2) \] Combining like terms gives: \[ 2x \ln(5) - x \ln(2) = -\ln(2) - 6 \ln(5) \] Factoring out \(x\): \[ x(2 \ln(5) - \ln(2)) = -\ln(2) - 6 \ln(5) \] Finally, isolate \(x\): \[ x = \frac{-\ln(2) - 6 \ln(5)}{2 \ln(5) - \ln(2)} \] Using a calculator to approximate this solution: Plug in the values for \(\ln(2)\) and \(\ln(5)\): \(\ln(2) \approx 0.693\) and \(\ln(5) \approx 1.609\), the computation will yield a decimal approximation for \(x\). Fun fact: Exponential equations like this one pop up in various fields, especially in interest calculations! You can find these equations anywhere, from finance to population growth models. Understanding how to manipulate and solve them can be incredibly powerful! For those looking to dive deeper, consider studying logarithmic properties further, like the change of base formula or how logarithms interact with exponential growth in real-world situations. These concepts are not only fascinating but crucial for many applications in science, economics, and data analysis!

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