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e^(2x + 3) - 8 = 0 x=???

ln(5 - 2x) = -6 x= ???

ln(ln x) = 1 x= ???

e^(ax) = Ce^(bx), where a is not = to b. x= ???

5 < ln x < 9


e2 - 3x > 4

2006-09-05 10:48:34 · 5 answers · asked by Anonymous in Science & Mathematics Mathematics

5 answers

Luekas is wrong on every one. I think it was just a cheap
way of making ponts.

All my answers agree with godmike. But I'd like to share
my method with you all on the 4th one

e^(ax)=Ce^(bx)
C=e^(ax)/e^(bx)
C=e^(ax-bx) taking ln of both sides
ax-bx=lnC
x(a-b)=lnC
x=lnC/(a-b)

2006-09-05 11:32:21 · answer #1 · answered by albert 5 · 0 1

Keep in mind that ln and e cancel each other out

e^(2x+3) - 8 = 0
e^(2x+3) = 8, take the ln of both sides
2x+3 = ln(8), solve for x
x = (ln(8)-3)/2 = -0.46 (appox)

ln(5-2x) = -6, take e of both sides
e^(ln(5-2x)) = e^(-6)
5-2x = e^(-6), solve for x
x = (5-e^(-6))/2 = 2.50 (appox)

ln(ln(x)) = 1, take e of both sides
e^ln(ln(x)) = e^1 = e
ln(x) = e, take e of both sides again
e^(ln(x)) = e^e
x = 15.15 (appox)

e^(ax) = Ce^(bx), take ln of both sides
ln(e^(ax)) = ln(Ce^(bx)) = ln(C) + ln(e^(bx))
ax = ln(C) + bx, solve for x
ax - bx = ln(C)
x(a-b) = ln(C)
x = ln(C)/(a-b)

5 < ln x < 9, take e of all sides
e^5 < e^ln(x) < e^9
148.41 < x < 8103.10 (approx)

e^2 - 3x > 4, solve for x
-3x > 4 - e^2
x < (e^2 - 4)/3
x < 1.12 (appox)

2006-09-05 11:04:49 · answer #2 · answered by godmike 2 · 2 0

Listen to Godmike.

2006-09-05 11:27:54 · answer #3 · answered by Rance D 5 · 1 0

x=5

x=8

x=2578

x=65

354>93

675566809653356>1

Albert I know. It was just supposed to be a small joke.

2006-09-05 10:50:58 · answer #4 · answered by Luekas 4 · 0 0

i started solving bt ii had so much more importent things to do

2006-09-05 10:56:27 · answer #5 · answered by Rami 5 · 0 0

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