Solve for x: Natural Log Help!

[tommy2st]

natural log helpp!

Homework Statement

ln(x+10)+ln(x+4)=2



2. The attempt at a solution

okay so i took it down to (-14+-(14^2 - 4(e^2 - 40))^1/2) / 2

which i broke down to 356 - 4e^2 inside the sqrt.

am i on the right track and if so Whats the next step?

 

[wukunlin]

looks ok, I'm not sure what you mean by next step though
x = (-14+-(14^2 - 4(e^2 - 40))^1/2) / 2 is your answer

 

[tommy2st]

looks ok, I'm not sure what you mean by next step though
x = (-14+-(14^2 - 4(e^2 - 40))^1/2) / 2 is your answer

it can be broken down further into something more simplified.

the example i have goes down from x^2 + 4x -(e^2 + 5)=0

to x= -2+-sqrt 9+e^2

but it doesn't show the steps used

 

[tommy2st]

i figured it out. the answer was -7+ (9+e^2)^1/2

 

[legendary_]

ln(x+10)+ln(x+4)=2
ln[(x+10)(x+4)] = 2
(x+10)(x+4) = e^2
x^2 + 14x + 40 = e^2
x^2 + 14x + 40-e^2 = 0
x = ( -14 +- sqrt((14^2)-4*1*(40-e^2)) ) / (2) [quadratic formula]
x = answer

 

[HallsofIvy]

ln(x+10)+ln(x+4)=2
ln[(x+10)(x+4)] = 2
(x+10)(x+4) = e^2
x^2 + 14x + 40 = e^2
x^2 + 14x + 40-e^2 = 0
x = ( -14 +- sqrt((14^2)-4*1*(40-e^2)) ) / (2) [quadratic formula]
x = answer

No. if you take the negative in the "+-", you get a number less than -10 so that both x+10 and x+ 4 are negative and you cannot take the logarithm.

Only ( -14 + sqrt((14^2)-4*1*(40-e^2)) ) / (2) satifies the original equation.