Solve for x: Natural Log Help!

AI Thread Summary
The discussion focuses on solving the equation ln(x+10) + ln(x+4) = 2. The user simplifies the equation to (x+10)(x+4) = e^2, leading to the quadratic x^2 + 14x + 40 - e^2 = 0. The quadratic formula is applied to find x, resulting in x = (-14 ± sqrt(14^2 - 4(40 - e^2))) / 2. It is emphasized that only the positive root satisfies the logarithmic conditions, as the negative root would yield invalid logarithmic arguments. The solution process highlights the importance of ensuring that the values of x keep the arguments of the logarithms positive.
tommy2st
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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?
 
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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
 


wukunlin said:
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
 


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


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
 


legendary_ said:
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.
 
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