Chen Mann
12/16/2023 · High School
(1 point) Find all solutions \( x \) to each of the following equations. Separate multiple answers with commas. Use no solution or none if there are no solutions. Your answer may be a formula, but it should not have any literal \( x \) in it. \( 3 e^{B x}=5 \) has solutions \( x=\square \) \( 10 e^{B x}=7 e^{B x}+5 \) has solutions \( x=\square \) \( 3 e^{B x}=5 e^{C x} \) has solutions \( x=\square \) \( 7 e^{B x}=4 e^{B x}+5 e^{C x} \) has solutions \( x=\square \) \( 7 e^{B x}=4 e^{B x}+20 e^{C x} \) has solutions \( x=\square \) \( \frac{e^{B x}}{e^{B x}+5 e^{C x}}=\frac{4}{7} \) has solutions \( x=\square \)
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Tutor-Verified Answer
Step-by-step Solution
To solve the equations, we will isolate \( x \) in each case. Here are the solutions for each equation:
1. **Equation:** \( 3 e^{B x} = 5 \)
**Solution:**
\[
e^{B x} = \frac{5}{3} \implies B x = \ln\left(\frac{5}{3}\right) \implies x = \frac{1}{B} \ln\left(\frac{5}{3}\right)
\]
**Answer:** \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
2. **Equation:** \( 10 e^{B x} = 7 e^{B x} + 5 \)
**Solution:**
\[
10 e^{B x} - 7 e^{B x} = 5 \implies 3 e^{B x} = 5 \implies e^{B x} = \frac{5}{3} \implies B x = \ln\left(\frac{5}{3}\right) \implies x = \frac{1}{B} \ln\left(\frac{5}{3}\right)
\]
**Answer:** \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
3. **Equation:** \( 3 e^{B x} = 5 e^{C x} \)
**Solution:**
\[
\frac{3}{5} = \frac{e^{C x}}{e^{B x}} \implies \frac{3}{5} = e^{(C - B)x} \implies (C - B)x = \ln\left(\frac{3}{5}\right) \implies x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right)
\]
**Answer:** \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
4. **Equation:** \( 7 e^{B x} = 4 e^{B x} + 5 e^{C x} \)
**Solution:**
\[
7 e^{B x} - 4 e^{B x} = 5 e^{C x} \implies 3 e^{B x} = 5 e^{C x} \implies \frac{3}{5} = \frac{e^{C x}}{e^{B x}} \implies \frac{3}{5} = e^{(C - B)x} \implies (C - B)x = \ln\left(\frac{3}{5}\right) \implies x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right)
\]
**Answer:** \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
5. **Equation:** \( 7 e^{B x} = 4 e^{B x} + 20 e^{C x} \)
**Solution:**
\[
7 e^{B x} - 4 e^{B x} = 20 e^{C x} \implies 3 e^{B x} = 20 e^{C x} \implies \frac{3}{20} = \frac{e^{C x}}{e^{B x}} \implies \frac{3}{20} = e^{(C - B)x} \implies (C - B)x = \ln\left(\frac{3}{20}\right) \implies x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right)
\]
**Answer:** \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
6. **Equation:** \( \frac{e^{B x}}{e^{B x} + 5 e^{C x}} = \frac{4}{7} \)
**Solution:**
\[
7 e^{B x} = 4(e^{B x} + 5 e^{C x}) \implies 7 e^{B x} = 4 e^{B x} + 20 e^{C x} \implies 3 e^{B x} = 20 e^{C x} \implies \frac{3}{20} = \frac{e^{C x}}{e^{B x}} \implies (C - B)x = \ln\left(\frac{3}{20}\right) \implies x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right)
\]
**Answer:** \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
In summary, the solutions are:
1. \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
2. \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
3. \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
4. \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
5. \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
6. \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
Quick Answer
1. \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
2. \( x = \frac{1}{B} \ln\left(\frac{5}{3}\right) \)
3. \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
4. \( x = \frac{1}{C - B} \ln\left(\frac{3}{5}\right) \)
5. \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
6. \( x = \frac{1}{C - B} \ln\left(\frac{3}{20}\right) \)
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