Cole Carter
02/07/2024 · Elementary 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. \( 3 e^{10 x}=5 e^{8 x} \) has solutions \( x=\square \) \( 7 e^{10 x}=4 e^{10 x}+5 e^{8 x} \) has solutions \( x=\square \) \( 7 e^{10 x}=4 e^{10 x}+20 e^{8 x} \) has solutions \( x=\square \) \( \frac{e^{10 x}}{e^{10 x}+5 e^{8 x}}=\frac{4}{7} \) has solutions \( x=\square \)
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Tutor-Verified Answer
Step-by-step Solution
To solve each of the equations, we will manipulate them algebraically.
1. **Equation:** \( 3 e^{10 x} = 5 e^{8 x} \)
Divide both sides by \( e^{8x} \) (assuming \( e^{8x} \neq 0 \)):
\[
3 e^{2x} = 5
\]
Now, divide both sides by 3:
\[
e^{2x} = \frac{5}{3}
\]
Take the natural logarithm of both sides:
\[
2x = \ln\left(\frac{5}{3}\right)
\]
Therefore:
\[
x = \frac{1}{2} \ln\left(\frac{5}{3}\right)
\]
**Solution:** \( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \)
2. **Equation:** \( 7 e^{10 x} = 4 e^{10 x} + 5 e^{8 x} \)
Rearranging gives:
\[
7 e^{10 x} - 4 e^{10 x} = 5 e^{8 x}
\]
Simplifying:
\[
3 e^{10 x} = 5 e^{8 x}
\]
Divide both sides by \( e^{8x} \):
\[
3 e^{2x} = 5
\]
This is the same equation as the first one, so:
\[
x = \frac{1}{2} \ln\left(\frac{5}{3}\right)
\]
**Solution:** \( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \)
3. **Equation:** \( 7 e^{10 x} = 4 e^{10 x} + 20 e^{8 x} \)
Rearranging gives:
\[
7 e^{10 x} - 4 e^{10 x} = 20 e^{8 x}
\]
Simplifying:
\[
3 e^{10 x} = 20 e^{8 x}
\]
Divide both sides by \( e^{8x} \):
\[
3 e^{2x} = 20
\]
Divide both sides by 3:
\[
e^{2x} = \frac{20}{3}
\]
Take the natural logarithm of both sides:
\[
2x = \ln\left(\frac{20}{3}\right)
\]
Therefore:
\[
x = \frac{1}{2} \ln\left(\frac{20}{3}\right)
\]
**Solution:** \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \)
4. **Equation:** \( \frac{e^{10 x}}{e^{10 x} + 5 e^{8 x}} = \frac{4}{7} \)
Cross-multiplying gives:
\[
7 e^{10 x} = 4 (e^{10 x} + 5 e^{8 x})
\]
Expanding the right side:
\[
7 e^{10 x} = 4 e^{10 x} + 20 e^{8 x}
\]
Rearranging gives:
\[
7 e^{10 x} - 4 e^{10 x} = 20 e^{8 x}
\]
Simplifying:
\[
3 e^{10 x} = 20 e^{8 x}
\]
Divide both sides by \( e^{8x} \):
\[
3 e^{2x} = 20
\]
Divide both sides by 3:
\[
e^{2x} = \frac{20}{3}
\]
Take the natural logarithm of both sides:
\[
2x = \ln\left(\frac{20}{3}\right)
\]
Therefore:
\[
x = \frac{1}{2} \ln\left(\frac{20}{3}\right)
\]
**Solution:** \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \)
### Summary of Solutions:
- \( 3 e^{10 x} = 5 e^{8 x} \) has solutions \( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \)
- \( 7 e^{10 x} = 4 e^{10 x} + 5 e^{8 x} \) has solutions \( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \)
- \( 7 e^{10 x} = 4 e^{10 x} + 20 e^{8 x} \) has solutions \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \)
- \( \frac{e^{10 x}}{e^{10 x} + 5 e^{8 x}} = \frac{4}{7} \) has solutions \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \)
Quick Answer
\( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \), \( x = \frac{1}{2} \ln\left(\frac{5}{3}\right) \), \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \), \( x = \frac{1}{2} \ln\left(\frac{20}{3}\right) \)
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