what is the solution to 4 + 5e^{x + 2}=11?\no x = ln\\frac{7}{5}-2\no x = ln\\frac{7}{5}+2\no x = ln35…
what is the solution to 4 + 5e^{x + 2}=11?\no x = ln\\frac{7}{5}-2\no x = ln\\frac{7}{5}+2\no x = ln35 - 2\no x = ln35+2
Answer
Explanation:
Step1: Isolate the exponential term
Subtract 4 from both sides of the equation $4 + 5e^{x + 2}=11$. $5e^{x + 2}=11 - 4=7$.
Step2: Solve for the exponential part
Divide both sides by 5. $e^{x + 2}=\frac{7}{5}$.
Step3: Use natural - logarithm property
Take the natural logarithm of both sides. Since $\ln(e^{a})=a$, we have $\ln(e^{x + 2})=\ln(\frac{7}{5})$. $x + 2=\ln(\frac{7}{5})$.
Step4: Solve for x
Subtract 2 from both sides. $x=\ln(\frac{7}{5})-2$.
Answer:
$x=\ln\frac{7}{5}-2$