Question

Please help with question!!

Answer 1

`\textsf{Exact form}: \quad x=-7 +√(49+e^3)`

`\textsf{Decimal form}: \quad x=1.312\; \sf (3\:d.p.)`

Explanation:

Given equation:

`\ln x + \ln (x+14)=3`

`\textsf{Apply the product law} \quad \ln x + \ln y=\ln xy:`

`\implies \ln (x(x+14))=3`

`\implies \ln (x^2+14x)=3`

`\textsf{Apply the log law} \quad e^(\ln x)=x:`

`\implies e^(\ln (x^2+14x))=e^3`

`\implies x^2+14x=e^3`

Subtract e³ from both sides to form a quadratic equation:

`\implies x^2+14x-e^3=e^3-e^3`

`\implies x^2+14x-e^3=0`

Quadratic Formula

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

Therefore:

• a = 1
• b = 14
• c = -e³

Substitute these values into the quadratic formula and solve for x:

`\implies x=(-14 \pm √(14^2-4(1)(-e^3)) )/(2(1))`

`\implies x=(-14 \pm √(196+4e^3) )/(2)`

`\implies x=(-14 \pm √(4(49+e^3)) )/(2)`

`\implies x=(-14 \pm √(4)√(49+e^3) )/(2)`

`\implies x=(-14 \pm 2√(49+e^3) )/(2)`

`\implies x=-7 \pm √(49+e^3)`

Solutions:

`x=-7 +√(49+e^3), \quad x=-7 -√(49+e^3)`

Decimal form:

`x= 1.312, \quad x=-15.312`

As logs of negative numbers cannot be taken, the only solution is:

`x=-7 +√(49+e^3)\\\\x=1.312\;\sf (3\;d.p.)`

Answer 2

Given:

`\ln x+\ln (x+14)=3`

Use the formula:

`\ln ab=\ln a+\ln b`
`\begin{gathered} \ln x+\ln (x+14)=3 \\ \ln x(x+14)=3 \\ x(x+14)=e^3 \\ x^2+14x=20.085 \end{gathered}`

solve the equation :

`\begin{gathered} x^2+14x=20.085 \\ x^2+14x-20.085=0 \\ x=\frac{-14\pm\sqrt[]{14^2-4(1)(-20.085)}}{2} \\ x=\frac{-14\pm\sqrt[]{276.34}}{2} \\ x=(-14\pm16.623)/(2) \\ x=-7\pm8.312 \\ x=-7+8.312;x=-7-8.312 \\ x=1.312;x=-15.312 \end{gathered}`