Question

Hello, I need some help with this precalculus question for my homework, please HW Q8

Answer 1

we have the equation

`ln(x)+ln(x+8)=4`

Applying property of log

`lnx*(x+8)=4`
`\begin{gathered} e^4=x(x+8) \\ e^4=x^2+8x \\ x^2+8x-e^4=0 \end{gathered}`

Find out the exact solution

Solve the quadratic equation using the formula

a=1

b=8

c=-e^4

substitute

`x=(-8\pm√(8^2-4(1)(-e^4)))/(2(1))`
`x=(-8\pm√(64+4e^4))/(2)`

Simplify

`x=(-8\pm2√(16+e^4))/(2)`

therefore the exact solutions are

`\begin{gathered} x=-4+√(16+e^4) \\ x=-4-√(16+e^4) \end{gathered}`

The decimal approximation of the solutions are

`\begin{gathered} x=4.402 \\ x=-12.402 \end{gathered}`