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

[BackCountrySnow]

Q.6 a)
Q.7 a)

 

[sasha279]

Posting the questions might help??

 

[BackCountrySnow]

no i dont wanna.

 

[henry08]

lolz

 

[Daniel-08]

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]

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]

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]

6(a)

= -3pi/4 , 3pi/4

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

 

[minijumbuk]

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.

(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

ln x = 3 ln x =/= -1 as x > 0

True, but 1/e is positive.

 

[Azreil]

Oh fuck.

Oh well, thanks.

 

[Baggygreen408]

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