not sure is i got these right. plz halp. (1 Viewer)

Daniel-08

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ok 6.a) was x = 3pi/4, 9pi/4, -3pi/4, -9pi/4

for 7.a), you had to multiply all terms by ln x, making the function a quadratic

(ln x)^2 - 2(ln x) -3 = 0

sub in u = ln x

u^2 - 2u -3 =0

(u-3)(u+1) = 0

u = 3, u = 1

Therefore ln x =3, ln x =1

x = e^3, x = 1/e
 

Azreil

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Daniel-08 said:
ok 6.a) was x = 3pi/4, 9pi/4, -3pi/4, -9pi/4

for 7.a), you had to multiply all terms by ln x, making the function a quadratic

(ln x)^2 - 2(ln x) -3 = 0

sub in u = ln x

u^2 - 2u -3 =0

(u-3)(u+1) = 0

u = 3, u = 1

Therefore ln x =3, ln x =1

x = e^3, x = 1/e
Almost.

u=-1
lnx = -1 (no solutions)
therefore etc.
 

dwarven

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Daniel-08 said:
ok 6.a) was x = 3pi/4, 9pi/4, -3pi/4, -9pi/4

for 7.a), you had to multiply all terms by ln x, making the function a quadratic

(ln x)^2 - 2(ln x) -3 = 0

sub in u = ln x

u^2 - 2u -3 =0

(u-3)(u+1) = 0

u = 3, u = 1

Therefore ln x =3, ln x =1

x = e^3, x = 1/e
u = 3 u = -1
ln x = 3 ln x =/= -1 as x > 0
x = e^3
 

danz90

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6(a)

= -3pi/4 , 3pi/4

u dont include 9pi/4 in ur answer because thats out of the range -pi<x<pi
 

minijumbuk

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Azreil said:
Almost.

u=-1
lnx = -1 (no solutions)
therefore etc.
Press on your calculator ln (1/e).

I don't know how Daniel got the right answers, because he did his working wrong.
Daniel-08 said:
(ln x)^2 - 2(ln x) -3 = 0

sub in u = ln x

u^2 - 2u -3 =0

(u-3)(u+1) = 0

u = 3, u = -1

Therefore ln x =3, ln x = -1
x = e^3, x = 1/e

dwarven said:
ln x = 3 ln x =/= -1 as x > 0
True, but 1/e is positive.
 
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Baggygreen408

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Daniel-08 said:
ok 6.a) was x = 3pi/4, 9pi/4, -3pi/4, -9pi/4

for 7.a), you had to multiply all terms by ln x, making the function a quadratic

(ln x)^2 - 2(ln x) -3 = 0

sub in u = ln x

u^2 - 2u -3 =0

(u-3)(u+1) = 0

u = 3, u = 1

Therefore ln x =3, ln x =1

x = e^3, x = 1/e
the question, the firsteferenced 6) a), references a domain between - pi and pi
 

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