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In (x+8)+ In(x+4)=3. Solve for x?
it's impossible.
you can't solve it, even with a quadratic formula.
It needs to be written as
X= 12+-sqr(4^2) ( this is just an example
not X= 1.234
Please help me?
1 Answer
- 1 decade agoFavorite Answer
put left and right side to the power of e
e^(ln(x+8) + ln(x+4)) = e^3
e^(ln(x+8))*e^(ln(x+4)) = e^3
(x+8)*(x+4) = e^3
x^2 + 12x + 32 - e^3 = 0
x1,x2 = [-12 +- sqrt(144 - 128 + 4*e^3)] / 2
= - 6 +- [ 1/2 * sqrt(16 + 4*e^3) ]..........take 1/2 inside the sqrt (it's 1/4 in there :)
= -6 +- sqrt(4 + e^3)
x2 = -6 - sqrt(4 + e^3) = -10.9077018, which is not a solution, since x2 + 8 (x2 + 4, too) is negative and ln(x) is only defined on positive numbers.
x1 = -6 + sqrt(4 + e^3) = -1.0922982 is a solution, so this what you're looking for.
x = -6 + sqrt(4 + e^3)
hint:
e^3 is just a number ( = 20.0855), so if you're not allowed to write it in decimal form, don't - but still treat it like it's a number and part of "c" in the quadratic formula
ax^2 + bx + c = 0
