Question

Lösen Sie die folgenden Gleichungen fur x ∈ R:

a) ln( 1/ ³√ x) = 1 + 1/3 * ln √x⁵
x >0

b) 3 tan(x) - e^ix+ln(3) /sin(x)=2ix,   x ∈ ]−π/2, π/2[

Answer 1

ln( 1/ ³√ x) = 1 + 1/3 * ln √x⁵

-⅓•ln(x) =1+5/6 •ln(x)

- 7/6 •ln(x) =1

ln(x) = -6/7

x=e^-6/7

:-)

Answer 2

Hallo,

Aufgabe b)

$\frac{3 \tan (x)-e^{i x+\ln(3)}}{\sin (x)}=2$ i $x$

e^(ix) =cos(x)+i sin(x)

3(1/cos(x) -cos(x)/sin(x)) -3i=2ix

Realteil:

3(1/cos(x) -cos(x)/sin(x)) =0 |:3

(1/cos(x) -cos(x)/sin(x)) =0

sin(x)= cos²(x)

sin(x)= 1-sin²(x)

sin²(x) +sin(x)-1=0 ; z=sin(x)

z² +z-1=0

z1,2=-1/2 ±√(5/4)

------->

-1/2 +√(5/4) = sin(x) und ----->x≈38.2°

-1/2 -√(5/4) = sin(x) --->keine Lösung

Imaginärteil:

-3=2x ; x= -3/2