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Solve the following exponential equations, giving the final answers correct to 3 significant figures. i. \( 5^{2 x-1}=4^{300} \) \( x \approx \) ii. \( 2^{y+1}=\frac{10}{2^{y}} \). \( y \approx \)

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Mar 15,2025

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\( x \approx 129 \) and \( y \approx 0.349 \).

Solución

Solve the equation by following steps: - step0: Solve for \(y\): \(2^{y+1}=\frac{10}{2^{y}}\) - step1: Cross multiply: \(2^{y+1}\times 2^{y}=10\) - step2: Simplify the equation: \(2^{2y+1}=10\) - step3: Take the logarithm of both sides: \(\log_{2}{\left(2^{2y+1}\right)}=\log_{2}{\left(10\right)}\) - step4: Evaluate the logarithm: \(2y+1=\log_{2}{\left(10\right)}\) - step5: Move the constant to the right side: \(2y=\log_{2}{\left(10\right)}-1\) - step6: Divide both sides: \(\frac{2y}{2}=\frac{\log_{2}{\left(10\right)}-1}{2}\) - step7: Divide the numbers: \(y=\frac{\log_{2}{\left(10\right)}-1}{2}\) - step8: Simplify: \(y=\frac{\log_{2}{\left(5\right)}}{2}\) Solve the equation \( 5^{2x-1}=4^{300} \). Solve the equation by following steps: - step0: Solve for \(x\): \(5^{2x-1}=4^{300}\) - step1: Take the logarithm of both sides: \(\log_{5}{\left(5^{2x-1}\right)}=\log_{5}{\left(4^{300}\right)}\) - step2: Evaluate the logarithm: \(2x-1=\log_{5}{\left(4^{300}\right)}\) - step3: Simplify: \(2x-1=600\log_{5}{\left(2\right)}\) - step4: Move the constant to the right side: \(2x=600\log_{5}{\left(2\right)}+1\) - step5: Divide both sides: \(\frac{2x}{2}=\frac{600\log_{5}{\left(2\right)}+1}{2}\) - step6: Divide the numbers: \(x=\frac{600\log_{5}{\left(2\right)}+1}{2}\) Calculate or simplify the expression \( \log(5)/2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\log_{10}{\left(5\right)}}{2}\) Calculate or simplify the expression \( (600*\log(2)/\log(5)+1)/2 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(\frac{600\log_{10}{\left(2\right)}}{\log_{10}{\left(5\right)}}+1\right)}{2}\) - step1: Remove the parentheses: \(\frac{\frac{600\log_{10}{\left(2\right)}}{\log_{10}{\left(5\right)}}+1}{2}\) - step2: Use the logarithm base change rule: \(\frac{600\log_{5}{\left(2\right)}+1}{2}\) Let's solve the given exponential equations step by step. ### i. Solve \( 5^{2x-1} = 4^{300} \) 1. **Rewrite the equation**: \[ 5^{2x-1} = (2^2)^{300} = 2^{600} \] 2. **Take logarithm on both sides**: \[ (2x - 1) \log(5) = 600 \log(2) \] 3. **Isolate \( x \)**: \[ 2x - 1 = \frac{600 \log(2)}{\log(5)} \] \[ 2x = \frac{600 \log(2)}{\log(5)} + 1 \] \[ x = \frac{600 \log(2)}{2 \log(5)} + \frac{1}{2} \] 4. **Calculate the value**: \[ x \approx 129.703 \quad (\text{to 3 significant figures: } 129) \] ### ii. Solve \( 2^{y+1} = \frac{10}{2^{y}} \) 1. **Rewrite the equation**: \[ 2^{y+1} \cdot 2^{y} = 10 \] \[ 2^{2y+1} = 10 \] 2. **Take logarithm on both sides**: \[ (2y + 1) \log(2) = \log(10) \] 3. **Isolate \( y \)**: \[ 2y + 1 = \frac{\log(10)}{\log(2)} \] \[ 2y = \frac{\log(10)}{\log(2)} - 1 \] \[ y = \frac{\log(10)}{2 \log(2)} - \frac{1}{2} \] 4. **Calculate the value**: \[ y \approx 0.349 \quad (\text{to 3 significant figures: } 0.349) \] ### Final Answers: - \( x \approx 129 \) - \( y \approx 0.349 \)

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The Deep Dive

To solve the first equation \( 5^{2x-1} = 4^{300} \), we can take logarithms on both sides. 1. Taking logarithm: \[ \log(5^{2x-1}) = \log(4^{300}) \] 2. Applying the power rule: \[ (2x - 1) \log(5) = 300 \log(4) \] 3. Solving for \( x \): \[ 2x - 1 = \frac{300 \log(4)}{\log(5)} \] \[ 2x = \frac{300 \log(4)}{\log(5)} + 1 \] \[ x = \frac{1}{2} \left( \frac{300 \log(4)}{\log(5)} + 1 \right) \] Calculating \( x \) gives approximately \( x \approx 121.363 \), so to three significant figures, \( x \approx 121 \). For the second equation \( 2^{y+1} = \frac{10}{2^{y}} \): 1. Multiplying both sides by \( 2^{y} \): \[ 2^{y+1} \cdot 2^{y} = 10 \] 2. Simplifying: \[ 2^{2y + 1} = 10 \] 3. Taking logarithm: \[ (2y + 1) \log(2) = \log(10) \] 4. Solving for \( y \): \[ 2y + 1 = \frac{\log(10)}{\log(2)} \] \[ 2y = \frac{\log(10)}{\log(2)} - 1 \] \[ y = \frac{1}{2} \left( \frac{\log(10)}{\log(2)} - 1 \right) \] Calculating \( y \) gives approximately \( y \approx 0.5 \), so to three significant figures, \( y \approx 0.500 \).

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