Question

hw07a fall25. problem 4
(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.
3e^{bx}=5 has solutions x =
10e^{bx}=7e^{bx}+5 has solutions x =
3e^{bx}=5e^{cx} has solutions x =
7e^{bx}=4e^{bx}+5e^{cx} has solutions x =
7e^{bx}=4e^{bx}+20e^{cx} has solutions x =
\frac{e^{bx}}{e^{bx}+5e^{cx}}=\frac{4}{7} has solutions x =

Explanation:

Step1: Solve $3e^{Bx}=5$

Divide both sides by 3: $e^{Bx}=\frac{5}{3}$. Then take the natural - logarithm of both sides: $Bx = \ln(\frac{5}{3})$, so $x=\frac{\ln(\frac{5}{3})}{B}$ ($B
eq0$).

Step2: Solve $10e^{Bx}=7e^{Bx}+5$

Subtract $7e^{Bx}$ from both sides: $10e^{Bx}-7e^{Bx}=5$, which simplifies to $3e^{Bx}=5$. Then follow the same steps as in Step 1, and $x = \frac{\ln(\frac{5}{3})}{B}$ ($B
eq0$).

Step3: Solve $3e^{Bx}=5e^{Cx}$

Take the natural - logarithm of both sides: $\ln(3e^{Bx})=\ln(5e^{Cx})$. Using the property $\ln(ab)=\ln(a)+\ln(b)$, we get $\ln(3)+Bx=\ln(5)+Cx$. Rearrange terms: $Bx - Cx=\ln(5)-\ln(3)$. Factor out $x$: $x(B - C)=\ln(\frac{5}{3})$, so $x=\frac{\ln(\frac{5}{3})}{B - C}$ ($B
eq C$).

Step4: Solve $7e^{Bx}=4e^{Bx}+5e^{Cx}$

Subtract $4e^{Bx}$ from both sides: $3e^{Bx}=5e^{Cx}$. Then follow the same steps as in Step 3, and $x=\frac{\ln(\frac{5}{3})}{B - C}$ ($B
eq C$).

Step5: Solve $7e^{Bx}=4e^{Bx}+20e^{Cx}$

Subtract $4e^{Bx}$ from both sides: $3e^{Bx}=20e^{Cx}$. Take the natural - logarithm of both sides: $\ln(3)+Bx=\ln(20)+Cx$. Rearrange terms: $Bx - Cx=\ln(20)-\ln(3)$. Factor out $x$: $x(B - C)=\ln(\frac{20}{3})$, so $x=\frac{\ln(\frac{20}{3})}{B - C}$ ($B
eq C$).

Step6: Solve $\frac{e^{Bx}}{e^{Bx}+5e^{Cx}}=\frac{4}{7}$

Cross - multiply: $7e^{Bx}=4(e^{Bx}+5e^{Cx})$. Expand the right - hand side: $7e^{Bx}=4e^{Bx}+20e^{Cx}$. Then follow the same steps as in Step 5, and $x=\frac{\ln(\frac{20}{3})}{B - C}$ ($B
eq C$).

Answer:

$x=\frac{\ln(\frac{5}{3})}{B}$ ($B
eq0$); $x=\frac{\ln(\frac{5}{3})}{B}$ ($B
eq0$); $x=\frac{\ln(\frac{5}{3})}{B - C}$ ($B
eq C$); $x=\frac{\ln(\frac{5}{3})}{B - C}$ ($B
eq C$); $x=\frac{\ln(\frac{20}{3})}{B - C}$ ($B
eq C$); $x=\frac{\ln(\frac{20}{3})}{B - C}$ ($B
eq C$)